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second edition
LINEAR ALGEBRA
AND
MATRIX THEORY
Evar D. Nering
second
edition
Wiley
Ct Ht
ABOUT THE AUTHOR
EVAR D. NERING is Professor of
Mathematics and Chairman of the
Department at Arizona State Uni
versity, where he has taught
since 1960. Prior to that, he was
Assistant Professor of Mathe
matics at the University of Minne
sota (19481956), and Associate
Professor at the University of Ari
zona (19561960). He received his
A.B. and A.M., both in Mathe
matics, from Indiana University.
In 1947, he earned an A.M. from
Princeton University, and re
ceived his Ph.D. from the same
university in 1948. He worked as
a mathematician with Goodyear
Aircraft Corporation from 1953 to
1954.
During the summers of 1958, 1960,
and 1962, Dr. Nering was a visit
ing lecturer at the University of
Colorado. During the summer of
1961, he served as a Research
Mathematician at the Mathe
matics Research Center of the
University of Wisconsin.
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Linear Algebra and
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second edition
Evar D* Nering
Professor of Mathematics
Arizona State University
®
John Wiley & Sons, Inc.
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Copyright © 1963, 1970 by John Wiley & Sons, Inc.
AIJ rights reserved. No part of this book may be
reproduced by any means, nor transmitted, nor trans
lated into a machine language without the written
permission of the publisher.
Library of Congress Catalog Card Number: 769(646
SBN 471 63178 7 y^
Printed in the United States of America
10 987654321
Preface to
first edition
The underlying spirit of this treatment of the theory of matrices is that of a
concept and its representation. For example, the abstract concept of an
integer is the same for all cultures and presumably should be the same to a
being from another planet. But the various symbols we write down and
carelessly refer to as "numbers" are really only representations of the
abstract numbers. These representations should be called "numerals" and
we should not confuse a numeral with the number it represents. Numerals
of different types are the inventions of various cultures and individuals, and
the superiority of one system of numerals over another lies in the ease with
which they can be manipulated and the insight they give us into the nature
of the numbers they represent.
We happen to use numerals to represent things other than numbers. For
example, we put numerals (not numbers) on the backs of football players
to represent and identify them. This does not attribute to the football
players any of the properties of the corresponding numbers, and the usual
operations of arithmetic have no meaning in this context. No one would
think of adding the halfback, 20, to the fullback, 40, to obtain the guard, 60.
Matrices are used to represent various concepts with a wide variety of
different properties. To cover these possibilities a number of different
manipulations with matrices are introduced. In each situation the appro
priate manipulations that should be performed on a matrix or a set of
matrices depend critically on the concepts represented. The student who
learns the formalisms of matrix "arithmetic" without learning the under
lying concepts is in serious danger of performing operations which have no
meaning for the problem at hand.
The formal manipulations with matrices are relatively easy to learn, and
few students have any difficulty performing the operations accurately, if
somewhat slowly. If a course in matrix theory, however, places too much
emphasis on these formal steps, the student acquires an illfounded self
assurance. If he makes an error exactly analogous to adding halfbacks to
fullbacks to obtain guards, he usually considers this to be a minor error
(since each step was performed accurately) and does not appreciate how
serious a blunder he has made.
In even the simplest problems matrices can appear as representing several
V1 Preface to First Edition
different types of concepts. For example, it is typical to have a problem in
which some matrices represent vectors, some represent linear transforma
tions, and others represent changes of bases. This alone should make it
clear that an understanding of the things represented is essential to a meaning
ful manipulation of the representing symbols. In courses in vector analysis
and differential geometry many students have difficulty with the concepts
of co variant vectors and contra variant vectors. The troubles stem almost
entirely from a confusion between the representing symbols and the things
represented. As long as a student thinks of «tuples (the representing
symbols) as being identical with vectors (the things represented) he must
think there are two different "kinds" of vectors all jumbled together and he
sees no way to distinguish between them. There are, in fact, not two "kinds"
of vectors. There are two entirely distinct vector spaces ; their representations
happen to look alike, but they have to be manipulated differently.
Although the major emphasis in this treatment of matrix theory is on
concepts and proofs that some may consider abstract, full attention is given
to practical computational procedures. In different problems different
patterns of steps must be used, because the information desired is not the
same in all. Fortunately, although the patterns change, there are only a
few different types of steps. The computational techniques chosen here are
not only simple and effective, but require a variety of steps that is particularly
small. Because these same steps occur often, ample practice is provided
and the student should be able to obtain more than adequate confidence
in his skill.
A single pattern of discussion recurs regularly throughout the book.
First a concept is introduced. A coordinate system is chosen so that the
concept can be represented by «tuples or matrices. It is shown how the
representation of the concept must be changed if the coordinate system is
chosen in a different way. Finally, an effective computational procedure is
described by which a particular coordinate system can be chosen so that the
representing wtuples or matrices are either simple or yield significant in
formation. In this way computational skill is supported by an understanding
of the reasons for the various procedures and why they differ. Lack of this
understanding is the most serious single source of difficulty for many students
of matrix theory.
The material contained in the first five chapters is intended for use in a
onesemester, or onequarter, course in the theory of matrices. The topics
have been carefully selected and very few omissions should be made. The
sections recommended for omission are designated with asterisks. The part
of Section 4 of Chapter III following the proof of the HamiltonCayley
Theorem can also be omitted without harm. No other omission in the first
three chapters is recommended. For a onequarter course the following
Preface to First Edition vii
additional omissions are recommended: Chapter IV, Sections 4 and 5;
Chapter V, Sections 3, 4, 5, and 9.
Chapter V contains two parallel developments leading to roughly analogous
results. One, through Sections 1, 6, 7, 8, 10, 11, and 12, is rather formal
and is based on the properties of matrices; the other, through Sections
1, 3, 4, 5, 9, and 11, is more theoretical and is based on the properties of
linear functional . This latter material is, in turn, based on the material
in Sections 15 of Chapter IV. For this reason these omissions should not
be made in a random manner. Omissions can be made to accommodate a
onequarter course, or for other purposes, by carrying one line of development
to its completion and curtailing the other.
The exercises are an important part of this book. The computational
exercises in particular should be worked. They are intended to illustrate the
theory, and their cumulative effect is to provide assurance and understanding.
Numbers have been chosen so that arithmetic complications are minimized.
A student who understands what he is doing should not find them lengthy
or tedious. The theoretical problems are more difficult. Frequently they
are arranged in sequence so that one exercise leads naturally into the next.
Sometimes an exercise has been inserted mainly to provide a suggestive
context for the next exercise. In other places large steps have been broken
down into a sequence of smaller steps, and these steps arranged in a sequence
of exercises. For this reason a student may find it easier to work ten exercises
in a row than to work five of them by taking every other one. These exercises
are numerous and the student should not let himself become bogged down
by them. It is more important to keep on with the pace of development
and to obtain a picture of the subject as a whole than it is to work every
exercise.
The last chapter on selected applications of linear algebra contains a lot
of material in rather compact form. For a student who has been through
the first five chapters it should not be difficult, but it is not easy reading for
someone whose previous experience with linear algebra is in a substantially
different form. Many with experience in the applications of mathematics
will be unfamiliar with the emphasis on abstract and general methods. I
have had enough experience in fulltime industrial work and as a consultant
to know that these abstract ideas are fully as practical as any concrete
methods, and anyone who takes the trouble to familiarize himself with
these ideas will find that this is true. In all engineering problems and most
scientific problems it is necessary to deal with particular cases and with
particular numbers. This is usually, however, only the last phase of the
work. Initially, the problem must be dealt with in some generality until
some rather important decisions can be made (and this is by far the more
interesting and creative part of the work). At this stage methods which
viii Preface to First Edition
lead to understanding are to be preferred to methods which obscure under
standing in unnecessary formalisms.
This book is frankly a textbook and not a treatise, and I have not attempted
to give credits and references at each point. The material is a mosaic that
draws on many sources. I must, however, acknowledge my tremendous
debt to Professor Emil Artin, who was my principal source of mathematical
education and inspiration from my junior year through my Ph.D. The
viewpoint presented here, that matrices are mere representations of things
more basic, is as close to his viewpoint as it is possible for an admiring
student to come in adopting the ideas of his teacher. During my student
days the book presenting this material in a form most in harmony with
this viewpoint was Paul Halmos' elegant treatment in Finite Dimensional
Vector Spaces, the first edition. I was deeply impressed by this book, and
its influence on the organization of my text is evident.
Evar D. Nering
Tempe, Arizona
January, 1963
Preface to
second edition
This edition differs from the first primarily in the addition of new material.
Although there are significant changes, essentially the first edition remains
intact and embedded in this book. In effect, the material carried over from
the first edition does not depend logically on the new material. Therefore,
this first edition material can be used independently for a onesemester or
onequarter course. For such a onesemester course, the first edition usually
required a number of omissions as indicated in its preface. This omitted
material, together with the added material in the second edition, is suitable
for the second semester of a twosemester course.
The concept of duality receives considerably expanded treatment in this
second edition. Because of the aesthetic beauty of duality, it has long been
a favorite topic in abstract mathematics. I am convinced that a thorough
understanding of this concept also should be a standard tool of the applied
mathematician and others who wish to apply mathematics. Several sections
of the chapter concerning applications indicate how duality can be used.
For example, in Section 3 of Chapter V, the inner product can be used to
avoid introducing the concept of duality. This procedure is often followed
in elementary treatments of a variety of subjects because it permits doing
some things with a minimum of mathematical preparation. However, the
cost in loss of clarity is a heavy price to pay to avoid linear functionals.
Using the inner product to represent linear functionals in the vector space
overlays two different structures on the same space. This confuses concepts
that are similar but essentially different. The lack of understanding which
usually accompanies this shortcut makes facing a new context an unsure
undertaking. I think that the use of the inner product to allow the cheap
and early introduction of some manipulative techniques should be avoided.
It is far better to face the necessity of introducing linear functionals at the
earliest opportunity.
I have made a number of changes aimed at clarification and greater
precision. I am not an advocate of rigor for rigor's sake since it usually
adds nothing to understanding and is almost always dull. However, rigor
is not the same as precision, and algebra is a mathematical subject capable
of both precision and beauty. However, tradition has allowed several
situations to arise in which different words are used as synonyms, and all
IX
x Preface to Second Edition
are applied indiscriminately to concepts that are not quite identical. For
this reason, I have chosen to use "eigenvalue" and "characteristic value" to
denote nonidentical concepts; these terms are not synonyms in this text.
Similarly, I have drawn a distinction between dual transformations and
adjoint transformations.
Many people were kind enough to give me constructive comments and
observations resulting from their use of the first edition. All these comments
were seriously considered, and many resulted in the changes made in this
edition. In addition, Chandler Davis (Toronto), John H. Smith (Boston
College), John V. Ryff (University of Washington), and Peter R. Christopher
(Worcester Polytechnic) went through the first edition or a preliminary
version of the second edition in detail. Their advice and observations were
particularly valuable to me. To each and every one who helped me with
this second edition, I want to express my debt and appreciation.
Evar D. Nering
Tempe, Arizona
September, 1969
Contents
Introduction
I Vector spaces
1. Definitions 5
2. Linear Independence and Linear Dependence 10
3. Bases of Vector Spaces 15
4. Subspaces 20
II Linear transformations and matrices
1. Linear Transformations 27
2. Matrices 37
3. NonSingular Matrices 45
4. Change of Basis 50
5. Hermite Normal Form 53
6. Elementary Operations and Elementary Matrices 57
7. Linear Problems and Linear Equations 63
8. Other Applications of the Hermite Normal Form 68
9. Normal Forms 74
*10. Quotient Sets, Quotient Spaces 78
♦11. Hom(U, V) 83
III Determinants, eigenvalues, and similarity transformations 85
1. Permutations 86
2. Determinants 89
3. Cofactors 93
4. The HamiltonCayley Theorem 98
5. Eigenvalues and Eigenvectors 104
6. Some Numerical Examples 110
7. Similarity 113
*8. The Jordan Normal Form 118
IV Linear functionate, bilinear forms, quadratic forms 128
1. Linear Functionals 129
2. Duality 133
xi
xu Contents
3. Change of Basis 134
4. Annihilators 138
5. The Dual of a Linear Transformation 142
*6. Duality of Linear Transformations 145
*7. Direct Sums 147
8. Bilinear Forms 156
9. Quadratic Forms 160
10. The Normal Form 162
11. Real Quadratic Forms 168
12. Hermitian Forms 170
V Orthogonal and unitary transformations, normal matrices 175
1. Inner Products and Orthonormal Bases 176
*2. Complete Orthonormal Sets 182
3. The Representation of a Linear Functional by an Inner Product 186
4. The Adjoint Transformation 189
5. Orthogonal and Unitary Transformations 194
6. Orthogonal and Unitary Matrices 195
7. Superdiagonal Form 199
8. Normal Matrices 201
9. Normal Linear Transformations 203
10. Hermitian and Unitary Matrices 208
11. Real Vector Spaces 209
12. The Computational Processes 213
VI Selected applications of linear algebra 219
1. Vector Geometry 220
2. Finite Cones and Linear Inequalities 229
3. Linear Programming 239
4. Applications to Communication Theory 253
5. Vector Calculus 259
6. Spectral Decomposition of Linear Transformations 270
7. Systems of Linear Differential Equations 278
8. Small Oscillations of Mechanical Systems 284
9. Representations of Finite Groups by Matrices 292
10. Application of Representation Theory to Symmetric Mechanical
Systems 312
Appendix 319
Answers to selected exercises 325
Index 345
Introduction
Many of the most important applications of mathematics involve what are
known as linear methods. The idea of what is meant by a linear method
applied to a linear problem or a linear system is so important that it deserves
attention in its own right. We try to describe intuitively what is meant by
a linear system and then give some idea of the reasons for the importance of
the concept.
As an example, consider an electrical network. When the network receives
an input, an output from the network results. As is customary, we can con
sider combining two inputs by adding them and then putting their sum
through the system. This sum input will also produce an output. If the
output of the sum is the sum of the outputs the system is said to be additive.
We can also modify an input by changing its magnitude, by multiplying the
input by a constant factor. If the resulting output is also multiplied by the
same factor the system is said to be homogeneous. If the system is both
additive and homogeneous it is said to be linear.
The simplification in the analysis of a system that results from the knowl
edge, or the assumption, that the system is linear is enormous. If we know
the outputs for a collection of different inputs, we know the outputs for all
inputs that can be obtained by combining these inputs in various ways.
Suppose, for example, that we are considering all inputs that are periodic
functions of time with a given period. The theory of Fourier series tells us
that, under reasonable restrictions, these periodic functions can be represented
as sums of simple sine functions. Thus in analyzing the response of a linear
system to a periodic input it is sufficient to determine the response when the
input is a simple sine function.
So many of the problems that we encounter are assumed to be linear
problems and so many of the mathematical techniques developed are
L Introduction
inherently linear that a catalog of the possibilities would be a lengthy under
taking. Potential theory, the theory of heat, and the theory of small vibrations
of mechanical systems are examples of linear theories. In fact, it is not easy
to find brilliantly successful applications of mathematics to nonlinear
problems. In many applications the system is assumed to be linear even
though it is not. For example, the differential equation for a simple pendulum
is not linear since the restoring force is proportional to the sine of the dis
placement angle. We usually replace the sine of the angle by the angle in
radians to obtain a linear differential equation. For small angles this is a
good approximation, but the real justification is that linear methods are
available and easily applied.
In mathematics itself the operations of differentiation and integration are
linear. The linear differential equations studied in elementary courses are
linear in the sense intended here. In this case the unknown function is the
input and the differential operator is the system. Any physical problem that
is describable by a linear differential equation, or system of linear differential
equations, is also linear.
Matrix theory, vector analysis, Fourier series, Fourier transforms, and
Laplace transforms are examples of mathematical techniques which are
particularly suited for handling linear problems. In order for the linear
theory to apply to the linear problem it is necessary that what we have called
"inputs and outputs" and "linear systems" be representable within the theory.
In Chapter I we introduce the concept of vector space. The laws of combina
tion which will be defined for vector spaces are intended to make precise the
meaning of our vague statement, "combining these inputs in various ways."
Generally, one vector space will be used for the set of inputs and one vector
space will be used for the set of outputs. We also need something to represent
the "linear system," and for this purpose we introduce the concept of linear
transformation in Chapter II.
The next step is to introduce a practical method for performing the needed
calculations with vectors and linear transformations. We restrict ourselves
to the case in which the vector spaces are finite dimensional. Here it is
appropriate to represent vectors by ^tuples and to represent linear trans
formations by matrices. These representations are also introduced in
Chapters I and II.
Where the vector spaces are infinite dimensional other representations are
required. In some cases the vectors may be represented by infinite sequences,
or Fourier series, or Fourier transforms, or Laplace transforms. For
example, it is now common practice in electrical engineering to represent
inputs and outputs by Laplace transforms and to represent linear systems
by still other Laplace transforms called transfer functions.
The point is that the concepts of vector spaces and linear transformations
are common to all linear methods while matrix theory applies to only those
Introduction 3
linear problems that are finite dimensional. Thus it is of practical value to
discuss vector spaces and linear transformations as much as possible before
introducing the formalisms of wtuples and matrices. And, generally, proofs
that can be given without recourse to «tuples and matrices will be shorter,
simpler, and clearer.
The correspondences that can be set up between vectors and the //tuples
which represent them, and between linear transformations and the matrices
which represent them, are not unique. Therefore, we have to study the
totality of all possible ways to represent vectors and linear transformations
and the relations between these different representations. Each possible
correspondence is called a coordinization of the vector space, and the process
of changing from one correspondence to another is called a change of
coordinates.
Any property of the vector space or linear transformation which is
independent of any particular coordinatization is called an invariant or
geometric property. We are primarily interested in those properties of
vectors and linear transformations which are invariant, and, if we use a
coordinatization to establish such a property, we are faced with the problem
of showing that the conclusion does not depend on the particular coordina
tization being used. This is an additional reason for preferring proofs which
do not make use of any coordinatization.
On the other hand, if a property is known to be invariant, we are free to
choose any coordinate system we wish. In such a case it is desirable and
advantageous to select a coordinate system in which the problem we wish
to handle is particularly simple, or in which the properties we wish to establish
are clearly revealed. Chapter III and V are devoted to methods for selecting
these advantageous coordinate systems.
In Chapter IV we introduce ideas which allow us to define the concept
of distance in our vector spaces. This accounts for the principal differences
between the discussions in Chapters III and V. In Chapter III there is no
restriction on the coordinate systems which are permitted. In Chapter V the
only coordinate systems permitted are "Cartesian" ; that is, those in which
the theorem of Pythagoras holds. This additional restriction in permissible
coordinate systems means that it is more difficult to find advantageous
coordinate systems.
In addition to allowing us to introduce the concept of distance the material
in Chapter IV is of interest in its own right. There we study linear forms,
bilinear forms, and quadratic forms. They have application to a number of
important problems in physics, chemistry, and engineering. Here too,
coordinate systems are introduced to allow specific calculations, but proofs
given without reference to any coordinate systems are preferred.
Historically, the term "linear algebra" was originally applied to the study
of linear equations, bilinear and quadratic forms, and matrices, and their
4 Introduction
changes under a change of variables. With the more recent studies of Hilbert
spaces and other infinite dimensional vector spaces this approach has proved
to be inadequate. New techniques have been developed which depend less
on the choice or introduction of a coordinate system and not at all upon the
use of matrices. Fortunately, in most cases these techniques are simpler than
the older formalisms, and they are invariably clearer and more intuitive.
These newer techniques have long been known to the working mathe
matician, but until very recently a curious inertia has kept them out of books
on linear algebra at the introductory level.
These newer techniques are admittedly more abstract than the older
formalisms, but they are not more difficult. Also, we should not identify
the word "concrete" with the word "useful." Linear algebra in this more
abstract form is just as useful as in the more concrete form, and in most cases
it is easier to see how it should be applied. A problem must be understood,
formulated in mathematical terms, and analyzed before any meaningful
computation is possible. If numerical results are required, a computational
procedure must be devised to give the results with sufficient accuracy and
reasonable efficiency. All the steps to the point where numerical results are
considered are best carried out symbolically. Even though the notation and
terminology of matrix theory is well suited for computation, it is not necessar
ily the best notation for the preliminary work.
It is a curious fact that if we look at the work of an engineer applying
matrix theory we will seldom see any matrices at all. There will be symbols
standing for matrices, and these symbols will be carried through many steps
in reasoning and manipulation. Only occasionally or at the end will any
matrix be written out in full. This is so because the computational aspects
of matrices are burdensome and unnecessary during the early phases of work
on a problem. All we need is an algebra of rules for manipulating with them.
During this phase of the work it is better to use some concept closer to the
concept in the field of application and introduce matrices at the point where
practical calculations are needed.
An additional advantage in our studying linear algebra in its invariant
form is that there are important applications of linear algebra where the under
lying vector spaces are infinite dimensional. In these cases matrix theory
must be supplanted by other techniques. A case in point is quantum mechanics
which requires the use of Hilbert spaces. The exposition of linear algebra
given in this book provides an easy introduction to the study of such spaces.
In addition to our concern with the beauty and logic of linear algebra in
this form we are equally concerned with its utility. Although some hints
of the applicability of linear algebra are given along with its development,
Chapter VI is devoted to a discussion of some of the more interesting and
representative applications.
chapter
I
Vector spaces
In this chapter we introduce the concepts of a vector space and a basis for
that vector space. We assume that there is at least one basis with a finite
number of elements, and this assumption enables us to prove that the
vector space has a vast variety of different bases but that they all have the
same number of elements. This common number is called the dimension
of the vector space.
For each choice of a basis there is a onetoone correspondence between
the elements of the vector space and a set of objects we shall call ntuples.
A different choice for a basis will lead to a different correspondence between
the vectors and the ^tuples. We, regard the vectors as the fundamental
objects under consideration and the ntuples a representations of the vectors.
Thus, how a particular vector is represented depends on the choice of the
basis, and these representations are noninvariant. We call the ntuple the
coordinates of the vector it represents; each basis determines a coordinate
system.
We then introduce the concept of subspace of a vector space and develop
the algebra of subspaces. Under the assumption that the vector space is
finite dimensional, we prove that each subspace has a basis and that for
each basis of the subspace there is a basis of the vector space which includes
the basis of the subspace as a subset.
1 I Definitions
To deal with the concepts that are introduced we adopt some notational
conventions that are commonly used. We usually use sansserif italic letter
to denote sets.
a e S means a is an element of the set S.
a ^ S means a is not an element of the set S.
6 Vector Spaces  I
S c 7 means S is a subset of the set T.
S n T denotes the intersection of the sets S and T, the set of elements
in both S and T.
S u T denotes the union of the sets S and T, the set of elements in S or T.
T — S denotes the set of elements in T but not in S. In case T is the set
of all objects under consideration, we shall call T — S the
complement of S and denote it by CS.
S^i/u e M denotes a collection of sets indexed so that one set S^ is specified
for each element /u e M. M is called the index set.
n ^M$M denotes the intersection of all sets S^./j, e M.
^e/v^V denotes the union of all sets S^.ju, e M.
denotes the set with no elements, the empty set.
A set will often be specified by listing the elements in the set or by giving
a property which characterizes the elements of the set. In such cases we
use braces: {a, /?} is the set containing just the elements a and /S, {a  P}
is the set of all a with property P, {a^  fi e M} denotes the set of all a„
corresponding to [x in the index set M. We have such frequent use for the set
of all integers or a subset of the set of all integers as an index set that we
adopt a special convention for these cases, {aj denotes a set of elements
indexed by a subset of the set of integers. Usually the same index set is used
over and over. In such cases it is not necessary to repeat the specifications
of the index set and often designation of the index set will be omitted.
Where clarity requires it, the index set will be specified. We are careful to
distinguish between the set {aj and an element o^ of that set.
Definition. By a. field F we mean a nonempty set of elements with two laws
of combination, which we call addition and multiplication, satisfying the
following conditions :
Fl. To every pair of elements a, b e F there is associated a unique element,
called their sum, which we denote by a + b.
F2. Addition is associative; (a + b) + c = a + (b + c).
F3. There exists an element, which we denote by 0, such that a + = a
for all aeF.
FA. For each a e F there exists an element, which we denote by —a, such
that a + (—a) = 0. Following usual practice we write b + (—a) = b — a.
F5. Addition is commutative; a + b = b + a.
F6. To every pair of element a, b e F there is associated a unique element,
called their product, which we denote by ab, or a • b.
Fl. Multiplication is associative; (ab)c = a(bc).
FS. There exists an element different from 0, which we denote by 1,
such that a • 1 = a for all a e F.
1 I Definitions 7
F9. For each a e F, a ^ 0, there exists an element which we denote by
a 1 , such that a • a 1 = 1.
F10. Multiplicati6n is commutative: ab = ba.
F\\. Multiplication is distributive with respect to addition:
(a + b)c = ac + be.
The elements of F are called scalars, and will generally be denoted by lower
case Latin italic letters.
The rational numbers, real numbers, and complex numbers are familiar
and important examples of fields, but they do not exhaust the possibilities.
As a less familiar example, consider the set {0, 1} where addition is defined
by the rules: + 0=1 + 1=0, 0+1 = 1; and multiplication is defined
by the rules: 00 = 01=0, 11 = 1. This field has but two elements,
and there are other fields with finitely many elements.
We do not develop the various properties of abstract fields and we are
not concerned with any specific field other than the rational numbers, the
real numbers, and the complex numbers. We find it convenient and desirable
at the moment to leave the exact nature of the field of scalars unspecified
because much of the theory of vector spaces and matrices is valid for arbitrary
fields.
The student unacquainted with the theory of abstract fields will not be
handicapped for it will be sufficient to think of F as being one of the familiar
fields. All that matters is that we can perform the operations of addition and
subtraction, multiplication and division, in the usual way. Later we have to
restrict F to either the field of real numbers or the field of complex numbers
in order to obtain certain classical results; but we postpone that moment as
long as we can. At another point we have to make a very mild assumption,
that is, 1 + 1 5* 0, a condition that happens to be false in the example given
above. The student interested mainly in the properties of matrices with real
or complex coefficients should consider this to be no restriction.
Definition. A vector space V over F is a nonempty set of elements, called
vectors, with two laws of combination, called vector addition (or addition)
and scalar multiplication, satisfying the following conditions:
A\. To every pair of vectors a, $ e V there is associated a unique vector
in V called their sum, which we denote by <x + /?.
A2. Addition is associative; (a + /J) + y = a + (/? + y).
A3. There exists a vector, which we denote by 0, such that a + = a
for all a e V.
AA. For each ae V there exists an element, which we denote by —a,
such that a + ( — a) = 0.
A5. Addition is commutative; a + (3 = /? + a.
8 Vector Spaces  I
B\. To every scalar a e F and vector a e V, there is associated a unique
vector, called the product of a and a, which we denote by aa.
52. Scalar multiplication is associative : a(ba.) = (ab)tx..
B3. Scalar multiplication is distributive with respect to vector addition;
a(oc + /?) = aa + a/?.
54. Scalar multiplication is distributive with respect to scalar addition;
(a + b)ct. = aa + bca.
B5. 1 • a = a (where 1 e F).
We generally use lower case Greek letters to denote vectors. An exception
is the zero vector of A3. From a logical point of view we should not use the
same symbol "0 ' for both the zero scalar and the zero vector, but this practice
is rooted in a long tradition and it is not as confusing as it may seem at first.
The vector space axioms concerning addition alone have already appeared
in the definition of a field. A set of elements satisfying the first four axioms
is called a group. If the set of elements also satisfies A5 it is called a com
mutative group or abelian group. Thus both fields and vector spaces are
abelian groups under addition. The theory of groups is well developed and
our subsequent discussion would be greatly simplified if we were to assume
a prior knowledge of the theory of groups. We do not assume a prior
knowledge of the theory of groups; therefore, we have to develop some of
their elementary properties as we go along, although we do not stop to point
out that what was proved is properly a part of group theory. Except for
specific applications in Chapter VI we do no more than use the term "group"
to denote a set of elements satisfying these conditions.
First, we give some examples of vector spaces. Any notation other than
"F" for a field and "V" for a vector space is used consistently in the same
way throughout the rest of the book, and these examples serve as definitions
for these notations:
(1) Let F be any field and let V = P be the set of all polynomials in an
indeterminate x with coefficients in F. Vector addition is defined to be the
ordinary addition of polynomials, and multiplication is defined to be the
ordinary multiplication of a polynomial by an element of F.
(2) For any positive integer n, let P n be the set of all polynomials in x
with coefficients in F of degree < n — 1 , together with the zero polynomial.
The operations are defined as in Example (1).
(3) Let F = R, the field of real numbers, and take V to be the set of all
realvalued functions of a real variable. If/ and g are functions we define
vector addition and scalar multiplication by the rules
(f+g)(*)=f(*)+g(*)>
(af)(x) = a[f(x)].
1 I Definitions "
(4) Let F = R, and let V be the set of continuous realvalued functions
of a real variable. The operations are defined as in Example (3). The point
of this example is that it requires a theorem to show that A\ and B\ are
satisfied.
(5) Let F = R, and let V be the set of realvalued functions defined on
the interval [0, 1] and integrable over that interval. The operations are
defined as in Example (3). Again, the main point is to show that A\ and B\
are satisfied.
(6) Let F = R, and let V be the set of all real valued functions of a real
variable difterentiable at least m times (m a positive integer). The operations
are defined as in Example (3).
(7) Let F = R, and let V be the set of all realvalued functions difterentiable
d 2 y
at least twice and satisfying the differential equation — + y = 0.
(8) Let F = R, and let V = R n be the set of all real ordered wtuples,
a = (a lt a 2 , . . . , a n ) with a t e F. Vector addition and scalar multiplication
are. defined by the rules
(a lt ... ,a n ) + (b u ...,b n )=(a 1 + b 1 ,...,a n + b n ), ^ ^
a(a lt ... ,a n ) = (aa lt . . . , aa n ).
We call this vector space the ndimensional real coordinate space or the
real affine nspace. (The name "Euclidean nspace" is sometimes used, but
that term should be reserved for an affine wspace in which distance is defined.)
 (9) Let F n be the set of all wtuples of elements of F. Vector addition and
scalar multiplication are defined by the rules (1.2). We call this vector
space an ndimensional coordinate space.
An immediate consequence of the axioms defining a vector space is that
the zero vector, whose existence is asserted in A3, and the negative vector,
whose existence is asserted in A4, are unique. Specifically, suppose
satisfies A3 for all vectors in V and that for some a e V there is a 0' satisfying
the condition a + = a + 0' = a. Then 0' = 0' + = 0' + (a + (a)) =
(0' + a) + (a) = (a + 0') + (a) = a + (a) = 0. Notice that we
have proved not merely that the zero vector satisfying A3 for all a is unique;
we have proved that a vector satisfying the condition of A3 for some a. must
be the zero vector, which is a much stronger statement.
Also, suppose that to a given a there were two negatives, (—a) and
(—a)', satisfying the conditions of A4. Then (— a)' =. (— a)' + =
(a)' + a + (a) = (a) + a + (a)' = (a) + = (a). Both
these demonstrations used the commutative law, A5. Use of this axiom
could have been avoided, but the necessary argument would then have been
somewhat longer.
^L
10 Vector Spaces  I
Uniqueness enables us to prove that 0a = 0. (Here is an example of the V*y
seemingly ambiguous use of the symbol "0." The "0" on the left side is a Ly*
scalar while that on the right is a vector. However, no other interpretation 1
could be given the symbols and it proves convenient to conform to the
convention rather than introduce some other symbol for the zero vector.)
For each a e V, a = 1 • a = (1 + 0)a = 1 • a + • a = or+ • a. Thus *
• a = 0. In a similar manner we can show that (— l)a = — a oT+"X— T)a = / )
(1 — l)a = • a = 0. Since the negative vector is unique we see that \ 
(— l)a = — a. It also follows similarly that a • = 0.
EXERCISES
1 to 4. What theorems must be proved in each of the Examples (4), (5), (6), and
(7) to verify All To verify B\ ? (These axioms are usually the ones which require
most specific verification. For example, if we establish that the vector space
described in Example (3) satisfies all the axioms of a vector space, then A\ and B\
are the only ones that must be verified for Examples (4), (5), (6), and (7). Why?)
5. Let P + be the set of polynomials with real coefficients and positive constant
term. Is P + a vector space ? Why ?
6. Show that if aa =0 and a ¥=■ 0, then a = 0. {Hint: Use axiom F9 for fields.)
7. Show that if aa =0 and a^fl, then a = 0.
8. Show that the £ such that a + f = p is (uniquely) £ = p + (a).
9. Let a = (2, —5, 0, 1) and P = (—3, 3, 1, —1) be vectors in the coordinate
space R 4 . Determine
(a) a+p.
{b) *p.
(c) 3a.
{d) 2a + 3/3.
10. Show that any field can be considered to be a vector space over itself.
11. Show that the real numbers can be considered to be a vector space over the
rational numbers.
12. Show that the complex numbers can be considered to be a vector space over
the real numbers.
13. Prove the uniqueness of the zero vector and the uniqueness of the negative
of each vector without using the commutative law, A5.
2 I Linear Independence and Linear Dependence
Because of the associative law for vector addition, we can omit the
parentheses from expressions like a^ + (a 2 <x 2 + a 3 <x s ) = {^^ + a 2 a 2 ) +
« 3 a 3 and write them in the simpler form a^ + a 2 tx. 2 + a 3 a 3 = ^Li a * a i
It is clear that this convention can be extended to a sum of any number of
2  Linear Independence and Linear Dependence 11
such terms provided that only a finite number of coefficients are different
from zero. Thus, whenever we write an expression like ^ a^ (in which we
do not specify the range of summation), it will be assumed, tacitly if not
explicitly, that the expression contains only a finite number of nonzero
coefficients.
If /5 = ^t fl i a i» we sa y tnat $ 1S a linear combination of the o^. We also
say that /8 is linearly dependent on the a* if can be expressed as a linear
combination of the o^. An expression of the form 2» fl z a i = is called a
linear relation among the <x f . A relation with all a t = is called a trivial
linear relation; a relation in which at least one coefficient is nonzero is
called a nontrivial linear relation.
Definition. A set of vectors is said to be linearly dependent if there exists a
nontrivial linear relation among them. Otherwise, the set is said to be
linearly independent.
It should be noted that any set of vectors that includes the zero vector is
linearly dependent. A set consisting of exactly one nonzero vector is linearly
independent. For if aa = with a ^ 0, then a = 1 • a = (a 1 • a)a. =
ar\aa) — cr 1 • = 0. Notice also that the empty set is linearly independent.
It is clear that the concept of linear independence of a set would be mean
ingless if a vector from a set could occur arbitrarily often in a possible relation.
If a set of vectors is given, however, by itemizing the vectors in the set it is a
definite inconvenience to insist that all the vectors listed be distinct. The
burden of counting the number of times a vector can appear in a relation is
transferred to the index set. For each index in the index set, we require that
a linear relation contain but one term corresponding to that index. Similarly,
when we specify a set by itemizing the vectors in the set, we require that one and
only one vector be listed for each index in the index set. But we allow the
possibility that several indices may be used to identify the same vector. Thus
the set {<x. lt a 2 }, where a x = a 2 is linearly dependent, and any set with any
vector listed at least twice is linearly dependent. To be precise, the concept
of linear independence is a property of indexed sets and not a property of sets.
In the example given above, the relation a x — a 2 = involves two terms in
the indexed set {a*  i e {1, 2}} while the set {a l5 a 2 } actually contains only
one vector. We should refer to the linear dependence of an indexed set rather
than the linear dependence of a set. The conventional terminology, which
we are adopting, is inaccurate. This usage, however, is firmly rooted in
tradition and, once understood, it is a convenience and not a source of
difficulty. We speak of the linear dependence of a set, but the concept always
refers to an indexed set. For a linearly independent indexed set, no vector can
be listed twice; so in this case the inaccuracy of referring to a set rather than
an indexed set is unimportant.
12 Vector Spaces  I
The concept of linear dependence and independence is used in essentially
two ways. (1) If a set {aj of vectors is known to be linearly dependent,
there exists a nontrivial linear relation of the form ]£t a i Cf i — 0 (This
relation is not unique, but that is usually incidental.) There is at least one
nonzero coefficient; let a k be nonzero. Then a fc = ^ i ^ k {—af 1 a^)a. i ; that is
one of the vectors of the set {aj is a linear combination of the others. (2) If a
set {aj of vectors is known to be linearly independent and a linear relation
2i a^i = is obtained, we can conclude that all a t = 0. This seemingly
trivial observation is surprisingly useful.
In Example (1) the zero vector is the polynomial with all coefficients
equal to zero. Thus the set of monomials {1, x, x 2 , . . .} is a linearly in
dependent set. The set {l,x,x 2 ,x 2 + x + 1} is linearly dependent since
1 f x + x 2 — (x 2 + x + 1) = 0. In P n of Example (2), any n + 1 poly
nomials form a linearly dependent set.
In R 3 consider the vectors {a = (1, 1, 0), p = (1, 0, 1), y = (0, 1, 1),
b — (1, 1, 1)}. These four vectors are linearly dependent since a + /3 +
y — 2d = 0, yet any three of these four vectors are linearly independent.
Theorem 2.1. If a is linearly dependent on {&} and each & is linearly
dependent on {/,}, then a is linearly dependent on {y 3 }.
proof. From a = ^ bfii and & = ^ c^y^ it follows that a =
Theorem 2.2. A set of nonzero vectors {a l5 a 2 , . . .} is linearly dependent
if and only if some tx. k is a linear combination of the a, with j < k.
proof. Suppose the vectors {a 1? a 2 , . . .} are linearly dependent. Then
there is a nontrivial linear relation among them; 2* a i y i — 0. Since a
positive finite number of coefficients are nonzero, there is a last nonzero
coefficient a k . Furthermore, k > 2 since <x. x ^ 0. Thus a fc = — a^^ti^i*** =
The converse is obvious. D
For any subset A of V the set of all linear combinations of vectors in A
is called the set spanned by A, and we denote it by (A). We also say that A
spans (A). It is a part of this definition that A c (A). We also agree that the
empty set spans the set consisting of the zero vector alone. It is readily
apparent that if A c 8, then (A) c <8>.
In this notation Theorem 2.1 is equivalent to the statement: If A c <8>
and 8 c: <C>, then A c <C>.
Theorem 2.3. The set {aj of nonzero vectors is linearly independent if
and only if for each k, <x k $ <a l9 . . . , a^). (To follow our definitions exactly,
the set spanned by {a l5 . . . , a^} should be denoted by ({o^, . . . , a^}).
2  Linear Independence and Linear Dependence 13
We shall use the symbol <a ls . . . , a^) instead since it is simpler and there is
no danger of ambiguity.)
proof. This is merely Theorem 2.2 in contrapositive form and stated in
new notation. □ , r . „ . ,
<fc. c * ? / •
Theorem 2.4. IfB and C are any subsets such thato <= (C), then (B) <= (Q.
proof. Set A = (B) in Theorem 2.1. Then 8 c <C) implies that <B> =
A c <C>. D
Theorem 2.5. If cn k e A is dependent on the other vectors in A, then (A) =
(A  K}>.
proof. The assumption that a fc is dependent on A — {a fc } means that
Ac.(A {a fc }>. It then follows from Theorem 2.4 that (A) c (A  (aj).
The equality follows from the fact that the inclusion in the other direction
is evident. □
Theorem 2.6. For any set C, ((C)) = (C).
proof. Setting 6 = (C) in Theorem 2.4 we obtain ((C)) = (B) c (p.
Again, the inclusion in the other direction is obvious. □
Theorem 2.7. If a finite set A = {a l5 . . . , a n } spans V, then every linearly
independent set contains at most n elements.
proof. Let B = {/?!, jg 2 , . . . } be a linearly independent set. We shall
successively replace the a f by the &, obtaining at each step a new nelement
set that spans V. Thus, suppose that A k = {^ t , . . . , &., a fc+1 , . . . , a„} is an
nelement set that spans V. (Our starting point, the hypothesis of the
theorem, is the case k = 0.) Since A k spans V, fi k+1 is dependent on A k .
Thus the set {jS l5 . . . , )8 fc , ]8 fc+1 , a fc+1 , . . . , a n } is linearly dependent. In any
nontrivial relation that exists the nonzero coefficients cannot be confined
to the j8,, for they are linearly independent. Thus one of the 0^(1 > k) is
dependent on the others, and after reindexing {a fc+1 , . . . , a n } if necessary
we may assume that it is a fc+1 . By Theorem 2.5 the set A k+1 = {/5 X , . . . ,
Pk+i, a *+2, • • • , <*«} also spans V.
If there were more than n elements in B, we would in this manner arrive
at the spanning set A n = {^, . . . , p n }. But then the dependence of /S n+1 on
A n would contradict the assumed linear independence of B. Thus B contains
at most n elements. □
Theorem 2.7 is stated in slightly different forms in various books. The
essential feature of the proof is the stepbystep replacement of the vectors
in one set by the vectors in the other. The theorem is known as the Steinitz
replacement theorem.
14 Vector Spaces  I
EXERCISES
1. In the vector space P of Example (1) let p x (x) = x 2 + x + 1, p 2 (x) =
x 2 — x — 2,p 3 (x) = x 2 + x — l,/> 4 0*0 = x — 1. Determine whether or not the set
{pi( x ), Pi( x ), Pz{ x )> Pi( x )} is linearly independent. If the set is linearly dependent,
express one element as a linear combination of the others.
2. Determine ({p x (x), p 2 (x), p z (x), p^x)}), where the p { (x) are the same poly
nomials as those defined in Exercise 1 . (The set required is infinite, so that we
cannot list all its elements. What is required is a description; for example, "all
polynomials of a certain degree or less," "all polynomials with certain kinds of
coefficients," etc.)
3. A linearly independent set is said to be maximal if it is contained in no larger
linearly independent set. In this definition the emphasis is on the concept of set
inclusion and not on the number of elements in a set. In particular, the definition
allows the possibility that two different maximal linearly independent sets might
have different numbers of elements. Find all the maximal linearly independent
subsets of the set given in Exercise 1. How many elements are in each of them?
4. Show that no finite set spans P; that is, show that there is no maximal finite
linearly independent subset of P. Why are these two statements equivalent ?
5. In Example (2) for n = 4, find a spanning set for P 4 . Find a minimal spanning
set. Use Theorem 2.7 to show that no other spanning set has fewer elements.
6. In Example (1) or (2) show that {1, x + 1, x 2 + x + 1, x 3 + x 2 + x + 1,
x* + x 3 + x 2 + x + 1} is a linearly independent set.
7. In Example (1) show that the set of all polynomials divisible by x  1 cannot
span P.
8. Determine which of the following set in R 4 are linearly independent over R.
(a) {(1,1,0,1), (1,1,1,1), (2,2,1,2), (0,1,0,0)}.
(b) {(1,0,0, 1), (0,1,1,0), (1,0,1,0), (0,1,0,1)}.
(c) {(1,0,0, 1), (0,1,0,1), (0,0,1,1), (1,1,1,1)}.
9. Show that {e x = (1, 0, 0, . . . , 0), e 2 = (0, 1, 0, . . . , 0), . . . , e n = (0, 0, 0,
. . . , 1)} is linearly independent in F n over F.
10. In Exercise 11 of Section 1 it was shown that we may consider the real
numbers to be a vector space over the rational numbers. Show that {1, V2} is a
linearly independent set over the rationals. (This is equivalent to showing that
Vl is irrational.) Using this result show that {1, V2, Vj} is linearly independent.
11. Show that if one vector of a set is the zero vector, then the set is linearly
dependent.
12. Show that if an indexed set of vectors has one vector listed twice, the set is
linearly dependent.
13. Show that if a subset of S is linearly dependent, then S is linearly dependent.
14. Show that if a set S is linearly independent, then every subset of S is linearly
independent.
3  Bases of Vector Spaces 15
15. Show that if the set A = {a l5 . . . , a n } is linearly independent and {a l5 . . . ,
a n , /3} is linearly dependent, then /S is dependent on A.
16. Show that, if each of the vectors {ft,, p lf . . . , fi n } is a linear combination of
the vectors {«!,...,«„}, then {p , p x , . . . , P n } is linearly dependent.
3 I Bases of Vector Spaces
Definition. A linearly independent set spanning a vector space V is called a
basis or base (the plural is bases) of V.
If A = {a l5 a 2 , . . .} is a basis of V, by definition anaeV can be written
in the form a = J< «i a i The interesting thing abotit a basis, as distinct from
other spanning sets, is that the coefficients are uniquely determined by a.
For suppose that we also have a = J, ^oq. Upon subtraction we get the
linear relation Ji ( fl < — ^i) a i = ° since M is a Nearly independent
set, flj — bi = and ^ = Z>, for each /. A related fact is that a basis is a
particularly efficient spanning set, as we shall see.
In Example (1) the vectors {a, = x l  i = 0, 1 , . . .} form a basis. We have
already observed that this set is linearly independent, and it clearly spans
the space of all polynomials. The space P„ has a basis with a finite number of
elements; {1, x, z 2 , . . . , x n  x ).
The vector spaces in Examples (3), (4), (5), (6), and (7) do not have bases
with a finite nuftiber of elements.
In Example (8) every R n has a finite basis consisting of {a,  a 4 = (d u ,
<5 2 i, • • • » dm)} (Here 6 ti is the useful symbol known as the Kronecker delta.
By definition d i6 = if i ^j and d u = 1.)
Theorem 3.1. If a vector space has one basis with a finite number of
elements, then all other bases are finite and have the same number of elements.
proof. Let A be a basis with a finite number n of elements, and let 8 be
any other basis. Since A spans V and B is linearly independent, by Theorem
2.7 the number m of element in 6 must be at most n. This shows that 8 is
finite and m < n. But then the roles of A and 6 can be interchanged to
obtain the inequality in the other order so that m = n. □
A vector space with a finite basis is called a finite dimensional vector
space, and the number of elements in a basis is called the dimension of
the space. Theorem 3.1 says that the dimension of a finite dimensional vector
space is well defined. The vector space with just one element, the zero
vector, has one linearly independent subset, the empty set. The empty set
is also a spanning set and is therefore a basis of {0}. Thus {0} has dimension
zero. There are very interesting vector spaces with infinite bases ; for example,
P of Example (1). Moreover, many of the theorems and proofs we give are
also valid for infinite dimensional vector spaces. It is not our intention,
16 Vector Spaces  I
however, to deal with infinite dimensional vector spaces as such, and when
ever we speak of the dimension of a vector space without specifying whether
it is finite or infinite dimensional we mean that the dimension is finite.
Among the examples we have discussed so far, each P n and each R n is
/idimensional. We have already given at least one basis for each. There
are many others. The bases we have given happen to be conventional and
convenient choices.
Theorem 3.2. Any n + 1 vectors in an ndimensional vector space are
linearly dependent.
proof. Their independence would contradict Theorem 2.7. □
We have already seen that the four vectors {a = (1, 1, 0), = (1, 0, 1),
y = (0, 1, 1), d = (1, 1, 1)} form a linearly dependent set in R 3 . Since R 3
is 3dimensional we see that this must be expected for any set containing
at least four vectors from R 3 . The next theorem shows that each subset of
three is a basis.
Theorem 3.3. A set of n vectors in an ndimensional vector space V is a
basis if and only if it is linearly independent. < ^ u* ^ h JUa**^ U^^W </^v,w ;^Y
proof. The "only if" is part of the definition of a basis. Let A =
{<*!, . . . , a J be a linearly independent set and let a be any vector in V.
Since {a 1? . . . , a„, a} contains n + 1 elements it must be linearly dependent.
Any nontrivial relation that exists must contain a with a nonzero coefficient,
for if that coefficient were zero the relation would amount to a relation in A.
Thus a is dependent on A. Hence A spans V and is a basis. □
Theorem 3.4. A set of n vectors in an ndimensional vector space V is a
basis if and only if it spans V. «  ■• . * '*' 1 * Y*""*" ' ' w ' ^ ^^ * ^f >^ u '<
proof. The "only if" is part of the definition of a basis. If n vectors did
span V and were linearly dependent, then (by Theorem 2.5) a proper subset
would also span V, contrary to Th e o rem 2 .7. □ ijn, <Uh^
We see that a basis is a maximal linearly independent set and a minimal
spanning set. This idea is made explicit in the next two theorems.
Theorem 3.5. In a finite dimensional vector space, every spanning set
contains a basis.
proof. Let B be a set spanning V. If V = {0}, then <= 8 is a basis of
{0}. If V 5^ {0}, then 6 must contain at least one nonzero vector a x . We
now search for another vector in B which is not dependent on {%}. We
call this vector a 2 and search for another vector in B which is not dependent
on the linearly independent set {a l9 a 2 }. We continue in this way as long as
we can, but the process must terminate as we cannot find more than n
£«. (M~t 13*2. . J, *U Wv.^ ^ W" <r^** W< "^ " >^vw, j^ , ^
3  Bases of Vector Spaces 17
linearly independent vectors in 6. Thus suppose we have obtained the set
A = {a l5 . . . , a TO } with the property that every vector in B is linearly de
pendent on A. Then because of Theorem 2. 1 the set A must also span V and
it is a basis. □
To drop the assumption that the vector space is ^dimensional would
change the complexion of Theorem 3.5 entirely. As it stands the theorem
is interesting but minor, and not difficult to prove. Without this assumption
the theorem would assert that every vector space has a basis since every
vector space is spanned by itself. Discussion of such a theorem is beyond the
aims of this treatment of the subject of vector spaces.
Theorem 3.6. In a finite dimensional vector space any linearly independent
set of vectors can be extended to a basis.
proof. Let A = {a l5 . . . , aj be a basis of V, and let 8 = {/? l5 . . . , /5 m }
be a linearly independent set (m < n). The set {/? l5 . . . , ft m , a l9 . . . , a M }
spans V. If this set is linearly dependent (and it surely is if m > 0) then
some element is a linear combination of the preceding elements (Theorem
2.2). This element cannot be one of the /3/s for then B would be linearly
dependent. But then this u. t can be removed to obtain a smaller set spanning
V (Theorem 2.5). We continue in this way, discarding elements as long as
we have a linearly dependent spanning set. At no stage do we discard one of
the /3/s. Since our spanning set is finite this process must terminate with a
basis containing B as a subset. □
Theorem 3.6 is one of the most frequently used theorems in the book.
It is often used in the following way. A nonzero vector with a certain desired
property is selected. Since the vector is nonzero, the set consisting of that
vector alone is a linearly independent set. An application of Theorem 3.6
shows that there is a basis containing that vector. This is usually the first step
of a proof by induction in which a basis is obtained for which all the vectors
in the basis have the desired property.
Let A = {a l5 . . . , a„} be an arbitrary basis of V, a vector space of dimension
n over the field F. Let a be any vector in V. Since A is a spanning set a can be
represented as a linear combination of the form <x = 2"=i a t a » Since A is
linearly independent this representation is unique, that is, the coefficients a t
are uniquely determined by a (for the given basis A). On the other hand,
for each ntuple (a t , . . . , a n ) there is a vector in V of the form ^Li a i a i
Thus there is a onetoone correspondence between the vectors in V and the
wtuples (fl l5 . . . , a n ) g F n .
If a = 2?=i fl i a i' th e sca l ar a i is called the ith coordinate of a, and a 4 a t 
is called the ith component of a. Generally, coordinates and components
depend on the choice of the entire basis and cannot be determined from
18 Vector Spaces  I
individual vectors in the basis. Because of the rather simple correspondence
between coordinates and components there is a tendency to confuse them and
to use both terms for both concepts. Since the intended meaning is usually
clear from context, this is seldom a source of difficulty.
If a = JjLx a^i corresponds to the «tuple (a lf . . . , a n ) and /? = 2?=x *t a i
corresponds to the ntuple (b lt . . . , b n ), then a. + /5 = 2?=i( a i + ^t) a i
corresponds to the «tuple (a x + b u . . . , a n + b n ). Also, act. = ^Li afir » a i
corresponds to the rctuple {aa x , . . . , aa n ). Thus the definitions of vector
addition and scalar multiplication among /ituples defined in Example (9)
correspond exactly to the corresponding operations in V among the vectors
which they represent. When two sets of objects can be put into a onetoone
correspondence which preserves all significant relations among their elements,
we say the two sets are isomorphic; that is, they have the same form. Using
this terminology, we can say that every vector space of dimension n over a
given field F is isomorphic to the wdimensional coordinate space F n . Two
sets which are isomorphic differ in details which are not related to their inter
nal structure. They are essentially the same. Furthermore, since two sets
isomorphic to a third are isomorphic to each other we see that all wdimen
sional vector spaces over the same field of scalars are isomorphic.
The set of «tuples together with the rules for addition and scalar multi
plication forms a vector spaceinitsown right. However, when a basis is chosen
in an abstract vector space V the correspondence described above establishes
an isomorphism between V and F n . In this context we consider F n to be a
representation of V. Because of the existence of this isomorphism a study
of vector spaces could be confined to a study of coordinate spaces. However,
the exact nature of the correspondence between V and F n depends upon the
choice of a basis in V. If another basis were chosen in V a correspondence
between the a g V and the «tuples would exist as before, but the correspond
ence would be quite different. We choose to regard the vector space V and the
vectors in V as the basic concepts and their representation by «tuples as a tool
for computation and convenience. There are two important benefits from
this viewpoint. Since we are free to choose the basis we can try to choose a
coordinatization for which the computations are particularly simple or for
which some fact that we wish to demonstrate is particularly evident. In
fact, the choice of a basis and the consequences of a change in basis is the
central theme of matrix theory. In addition, this distinction between a
vector and its representation removes the confusion that always occurs when
we define a vector as an ntuple and then use another «tuple to represent it.
Only the most elementary types of calculations can be carried out in the
abstract. Elaborate or complicated calculations usually require the intro
duction of a representing coordinate space. In particular, this will be re
quired extensively in the exercises in this text. But the introduction of
3  Bases of Vector Spaces 19
coordinates can result in confusions that are difficult to clarify without ex
tensive verbal description or awkward notation. Since we wish to avoid
cumbersome notation and keep descriptive material at a minimum in the
exercises, it is helpful to spend some time clarifying conventional notations
and circumlocutions that will appear in the exercises.
The introduction of a coordinate representation for V involves the selection
of a basis {<x x , . . . , a n } for V. With this choice a x is represented by (1, 0,
... , 0), a 2 is represented by (0, 1 , 0, . . . , 0), etc. While it may be necessary
to find a basis with certain desired properties the basis that is introduced at
first is arbitrary and serves only to express whatever problem we face in a
form suitable for computation. Accordingly, it is customary to suppress
specific reference to the basis given initially. In this context it is customary
to speak of "the vector {a x , a 2 , . . . , a„)" rather than "the vector a whose
representation with respect to the given basis {a l9 . . . , a n }is(a 1 , a 2 , . . . , a n )."
Such shortcuts may be disgracefully inexact, but they are so common that
we must learn how to interpret them.
For example, let V be a twodimensional vector space over R. Let A =
{a 1? a 2 } be the selected basis. If ^ = a x + a 2 an d j8 2 = — a i + a 2> then
8 = {/?!, j8 2 } is also a basis of V. With the convention discussed above we
would identify a x with (1,0), a 2 with (0, 1), /? x with (1, 1), and 8 2 with
(— 1, 1). Thus, we would refer to the basis 8 = {(1, 1), (—1, 1)}. Since
a i = IPi ~ £&> a i has the representation (\, —%) with respect to the basis 8.
If we are not careful we can end up by saying that "(1, 0) is represented by
EXERCISES
To show that a given set is a basis by direct appeal to the definition means
that we must show the set is linearly independent and that it spans V. In any
given situation, however, the task is very much simpler. Since V is ^dimensional
a proposed basis must have n elements. Whether this is the case can be told at a
glance. In view of Theorems 3.3 and 3.4 if a set has n elements, to show that it is a
basis it suffices to show either that it spans V or that it is linearly independent.
1. In R 3 show that {(1, 1,0), (1, 0, 1), (0, 1, 1)} is a basis by showing that it is
linearly independent.
2. Show that {(1,1, 0), (1,0, 1), (0, 1, 1)} is a basis by showing that <(1, 1, 0),
(1, 0, 1), (0, 1, 1)> contains (1, 0, 0), (0, 1,0) and (0, 0, 1). Why does this suffice?
3. In R 4 let A = {(1, 1, 0, 0), (0, 0, 1, 1), (1,0, 1, 0), (0, 1, 0, 1)} be a basis
(is it?) and let 8 ={(1,2, 1, 1), (0, 1,2, 1)} be a linearly independent set
(is it ?). Extend B to a basis of R 4 . (There are many ways to extend 8 to a basis.
It is intended here that the student carry out the steps of the proof of Theorem 3.6
for this particular case.)
20 Vector Spaces  I
4. Find a basis of R 4 containing the vector (1, 2, 3, 4). (This is another even
simpler application of the proof of Theorem 3.6. This, however, is one of the most
important applications of this theorem, to find a basis containing a particular
vector.)
5. Show that a maximal linearly independent set is a basis.
6. Show that a minimal spanning set is a basis.
4 I Subspaces
Definition. A subspace VV of a vector space V is a nonempty subset of V
which is itself a vector space with respect to the operations of addition and
scalar multiplication defined in V. In particular, the subspace must be a
vector space over the same field F.
The first problem that must be settled is the problem of determining the
conditions under which a subset W is in fact a subspace. It should be clear
that axioms A2, A5, B2, 53, 54, and 55 need not be checked as they are valid
in any subset of V. The most innocuous conditions seem to be Al and B\,
but it is precisely these conditions that must be checked. If B\ holds for a
nonempty subset VV, there is an a e VV so that 0<x = g VV. Also, for each
a e W, (— l) a = — a g W. Thus A3 and AA follow from B\ in any non
empty subset of a vector space and it is sufficient to check that VV is non
empty and closed under addition and scalar multiplication.
The two closure conditions can be combined into one statement: if
a, 8 g VV and a, b e F, then aa + bfi e VV. This may seem to be a small
change, but it is a very convenient form of the conditions. It is also equivalent
to the statement that all linear combinations of elements in VV are also in VV;
that is, <W> = VV. It follows directly from this statement that for any
subset A, (A) is a subspace. Thus, instead of speaking of the subset spanned
by A, we speak of the subspace spanned by A.
Every vector space V has V and the zero space {0} as subspaces. As a rule
we are interested in subspaces other than these and to distinguish them we
call the subspaces other than V and {0} proper subspaces. In addition, if
VV is a subspace we designate subspaces of VV other than VV and {0} as proper
subspaces of VV.
In Examples (1) and (2) we can take a fixed finite set {x x , x 2 , . . . , x m }
of elements of F and define VV to be the set of all polynomials such that
p{ Xl ) = p(x 2 ) = • • • = p(xj = 0. To show that VV is a subspace it is
sufficient to show that the sum of two polynomials which vanish at the
x t also vanishes at the x it and the product of a scalar and a polynomial
vanishing at the x t also vanishes at the x f . What is the situation in P n if
m> nl Similar subspaces can be defined in examples (3), (4), (5), (6),
and (7).
4  Subspaces 21
The space P m is a subspace of P, and also a subspace of P n for m <n.
In R n , for each m, < m <n, the set of all <x = (a u a 2 , . . . , a n ) such
that a x = a 2 = • • • = a m = is a subspace of R n . This subspace is proper
if < m < n.
Notice that the set of all ntuples of rational numbers is a subset of R n
and it is a vector space over the rational numbers, but it is not a subspace of
R n since it is not a vector space over the real numbers. Why ?
Theorem 4.1. The intersection of any collection of subspaces is a subspace.
proof. Let W^./biE M be an indexed collection of subspaces of V.
n jUe yviW /i is not empty since it contains 0. Let a, /S g n^VV,, and a,beF.
Then a, (3 e W^ for each ju e M. Since W M is a subspace aa + Z>/5 e W M for
each [is hA, and hence aa + Z>/5 e n^^W^. Thus H^^W^ is a subspace. □
Let A be any subset of V, not necessarily a subspace. There exist subspaces
W^ c: V which contain A; in fact, V is one of them. The intersection
^aczvv//^ of a ll such subspaces is a subspace containing A. It is the
smallest subspace containing A.
Theorem 4.2. For any A c V, n Acvv ^W /< = <A>; ?/?a? w, ?Ae smallest
subspace containing A is exactly the subspace spanned by A.
proof. Since H^^W^ is a subspace containing A, it contains all linear
combinations of elements of A. Thus (A) <= n AcW/4 W M . On the other
hand (A) is a subspace containing A, that is, (A) is one of the W M and hence
^w^ c <A>. Thus n AcWfl W^ = (A), n
W x + W 2 is defined to be the set of all vectors of the form a x + <x 2 where
a x G W x and a 2 6 W 2 .
Theorem 4.3. IfW x and W 2 are subspaces ofV, then W x + W 2 is a subspace
ofV.
proof. If a = ai + a 2 e W x + W 2 , = & + £ 2 e W x + W 2 , and a,
Z> e F, then act. + bfi = fl(a x + a 2 ) + 6(/?x + &) = (aaj + 6/^) + (aa 2 +
Z>/5 2 ) g W x + W 2 . Thus W x + W 2 is a subspace. □
Theorem 4.4. W x + W 2 /s //?e smallest subspace containing both W x and
W 2 ; f/zotf w, W x + VV 2 = (Wx U W 2 >. //"Ax spans W x and A 2 j/?a/!j W 2 , ^e«
A x u A 2 spans W x + W 2 .
proof. Since e W x , W 2 c W x + W 2 . Similarly, ^ c W x + VV 2 .
Since W x + VV 2 is a subspace containing Wx U W 2 , (W x u W 2 > c W x + W 2 .
For any a e W x + W 2 , a can be written in the form a = a x + a 2 where
a x g W x and a 2 g VV 2 . Then a x G W x <= (Wx u W 2 > and a 2 g W 2 cz (Wx U
W 2 >. Since (W 1 u W 2 > is a subspace, a = a x + a 2 e (Wx U W 2 >. Thus
Wx + W 2 = (Wx u W 2 >.
22 Vector Spaces  I
The second part of the theorem now follows directly. W t = (A x ) <=
(A x U Aj> and W 2 = (A 2 ) c (A x U A 2 ) so that W 1 u W 2 c (^ u A 2 > <=
<W! u W 2 >, and hence <W X U W 2 > = <A t u A 2 ). a
Theorem 4.5. A subspace W of an ndimensional vector space V is a finite
dimensional vector space of dimension m < n.
proof. If YV = {0}, then W is 0dimensional. Otherwise, there is a non
zero vector o^ e W. If (o^) = W, Wis 1dimensional. Otherwise, there is an
a 2 ^ (ax) in W. We continue in this fashion as long as possible. Suppose we
have obtained the linearly independent set {a l9 . . . , cc k } and that it does not
span W. Then there exists an a fc+1 e W, a fc+1 ^ (a l5 . . . , a fc ). In a linear
relation of the form 2*ii a^ = we could not have a k+1 ^ for then
a fc+i e <a l9 . . . , a t >. But then the relation reduces to the form JjLi «*<*» = 0.
Since {a x , . . . , aj is linearly independent, all a* = 0. Thus {a x , . . . , a fc , a^+j}
is linearly independent. In general, any linearly independent set in W that
does not span W can be expanded into a larger linearly independent set in W.
This process cannot go on indefinitely for in that event we would obtain more
than n linearly independent vectors in V. Thus there exists an m such that
<a 1} . . . , a TO > = W. It is clear that m < n. □
Theorem 4.6. Given any subspace W of dimension m in an ndimensional
vector space V, there exists a basis {a l5 . . . , a w , a TO+1 , . . . , a n } of V such that
{a x , . . . , oi m } is a basis of W.
proof. By the previous theorem we see that W has a basis {a x , . . . , a TO }.
This set is also linearly independent when considered in V, and hence by
Theorem 3.6 it can be extended to a basis of V. n
Theorem 4.7. If two subspaces U and Wofa vector space V have the same
finite dimension and U cz W, then U = W.
proof. By the previous theorem there exists a basis of U which can be
extended to a basis of W. But since dim U — dim W, the basis of W can
have no more elements than does the basis of U. This means a basis of
U is also a basis of W; that is, U = W. □
Theorem 4.8. If W x and W 2 are any two subspaces of a finite dimensional
vector space V, then dim (W t + W 2 ) = dim W 1 + dim VV 2  dim (W x n W 2 ).
proof. Let {a x , . . . , a r } be a basis of W 1 n VV 2 . This basis can be
extended to a basis {a x , . . . , a r , f} lt . . . , &} of W x and also to a basis
{ai, . . . , a r , y x , . . . , y t } of VV 2 . It is clear that {<x x , . . . , <x r , lf . . . , /? s , y lt
... ,y t } spans Wj + W 2 ; we wish to show that this set is linearly independent.
Suppose 2* a t at t + 2* M; + 2* c fcn = is a linear relation. Then
2< a &i + Z> Mi = — 2fc c k7k The left side is m w i and tne ri 8 nt side is in
VV 2 , and hence both are in W x n VV 2 . Each side is then expressible as a
linear combination of the {<xj. Since any representation of an element as a
linear combination of the {ai, . . . , a r , p it . . . , &} is unique, this means that
4  Subspaces 23
bj = for ally. By a symmetric argument we see that all c k = 0. Finally,
this means that J, a^ = from which it follows that all a t = 0. This
shows that the spanning set {a l5 . . . , a r , ft, . . . , ft, y x , . . . , 7t } is linearly
independent and a basis of W x + W 2 . Thus dim (Wj + VV 2 ) = r + 5 + / =
(r + 5) + +  r = dim W x + dim W 2  dim (W x n W 2 ). n
As an example, consider in R 3 the subspaces W x = <(1, 0, 2), (1, 2, 2))
and W 2 = <(1, 1, 0), (0, 1, 1)). Both subspaces are of dimension 2. Since
Wi <= W x + VV 2 cz R3 we see t h a t 2 < dim (W x + W 2 ) < 3. Because of
Theorem 4.8 this implies that 1 < dim (W x n W 2 ) < 2. In more familiar
terms, W x and W 2 are planes in a 3dimensional space. Since both planes
contain the origin, they do intersect. Their intersection is either a line or,
in case they coincide, a plane. The first problem is to find a basis for W x n W 2 .
Any a.eW 1 nW 2 must be expressible in the forms a = a(l, 0, 2) +
6(1, 2, 2) = c(l, 1, 0) + </(0, 1, 1). This leads to the three equations:
a + 6 = c
2b = c + d
2a + 2b= d.
These equations have the solutions b = —3a, c = —2a, d = —4a. Thus
a = a(l, 0, 2)  3a(l, 2, 2) = a(2, 6, 4). As a check we also have
a = 2a(l, 1,0) 4a(0, 1,1) = a(2, 6, 4). We have determined
that {(1, 3, 2)} is a basis of W t n W 2 . Also {(1, 3, 2), (1, 0, 2)} is a basis
of W x and {(1, 3, 2), (1,1, 0)} is a basis of W 2 .
We are all familiar with the theorem from solid geometry to the effect
that two nonparallel planes intersect in a line, and the example above is an
illustration of that theorem. In spaces of dimension higher than 3, how
ever, it is possible for two subspaces of dimension 2 to have but one point
in common. For example, in R 4 the subspaces W x = <(1 , 0, 0, 0), (0, 1 , 0, 0)>
and W 2 = <(0, 0, 1,0), (0,0,0,1)) are each 2dimensional and W x n
W 2 = {0}, W, + W 2 = R*.
Those cases in which dim (W x n VV 2 ) = deserve special mention. If
W x nW 2 = {0} we say that the sum W x + VV 2 is direct: W x + W 2 is a
direct sum of W x and W 2 . To indicate that a sum is direct we use the notation,
W x e W 2 . For a e W x W 2 there exist a x e W x and a 2 e W 2 such that
a = a x + a 2 . This much is true for any sum of two subspaces. If the sum is
direct, however, <x x and <x 2 are uniquely determined by a. For if a = a x + a 2 =
^ + a;, then a t  a( = a 2  a 2 . Since the left side is in W x and the
right side is in VV 2 , both are in W x n W 2 . But this means ol x — <x. x = and
oc 2 — <x 2 = 0; that is, the decomposition of a into a sum of an element in W x
plus an element in W 2 is unique. If V is the direct sum of W x and VV 2 , we say
that W x and VV 2 are complementary and that VV 2 is a complementary subspace
of W l5 or a complement of W^.
24 Vector Spaces  I
The notion of a direct sum can be extended to a sum of any finite number
of subspaces. The sum W x + • • • + W k is said to be direct if for each i,
W t n Q^i Wj) = {0}. If the sum of several subspaces is direct, we use the
notation W x © W 2 © • • • © W k . In this case, too, a e W x © • • ■ © W k
can be expressed uniquely in the form a = ^ a i5 a f e W^.
Theorem 4.9. If W is a subspace of V there exists a subspace W such that
V =w@w.
proof. Let {a l5 . . . , a m } be a basis of W. Extend this linearly inde
pendent set to a basis {a 1} . . . , a m , a TO+1 , . . . , a n } of V. Let W be the sub
space spanned by {<x TO+1 , . . . , <x„}. Clearly, W r\ W = {0} and the sum
V = W + W is direct. D
Thus every subspace of a finite dimensional vector space has a comple
mentary subspace. The complement is not unique, however. If for W there
exists a subspace W such that V = W © W, we say that W is a direct
summand of V.
Theorem 4.10. For a sum of several subspaces of a finite dimensional
vector space to be direct it is necessary and sufficient that dim (W x + • • • f
W k ) = dim W x + • • • + dim W k .
proof. This is an immediate consequence of Theorem 4.8 and the prin
ciple of mathematical induction. □
EXERCISES
1 . Let P be the space of all polynomials with real coefficients. Determine which
of the following subsets of P are subspaces.
(a){p(x)\p(l)=0}.
(b) {p{x)  constant term ofp(x) = 0}.
(c) {p(x)  degree ofp(x) = 3}.
{d) {p{x)  degree ofp(x) < 3}.
(Strictly speaking, the zero polynomial does not have a degree associated with it.
It is sometimes convenient to agree that the zero polynomial has degree less than
any integer, positive or negative. With this convention the zero polynomial is
included in the set described above, and it is not necessary to add a separate
comment to include it.)
(e) {p(x)  degree of p{x) is even} u {0}.
2. Determine which of the following subsets of R n are subspaces.
(a) {(«!, x 2 , . . . , x n )  x 1 = 0}.
(b) {(x lt x 2 ,..., x n )  x x > 0}.
(c) {(x x , x 2 , . . . , x n ) I x x + 2x 2 = 0}.
(d) {(»!, x 2 , . . . , x n )  x x + 2x 2 = 1}.
00 {(«i, *2, • • • , »«) I x l + 2x 2 ^ °}
(/) {(^ij x 2> ■ ■ ■ > x n) \ ™i < x i < Mf. i = 1, 2, . . . , n where the m^ and M t 
are constants}.
(g) {( x i, x z, • • • , x n) I *i = x 2 = ■ • • = ^n}
4  Subspaces 25
3. What is the essential difference between the condition used to define the
subset in (c) of Exercise 2 and the condition used in (d) ? Is the lack of a nonzero
constant term important in (c) ?
4. What is the essential difference between the condition used to define the
subset in (c) of Exercise 2 and the condition used in (e)7 What, in general, are the
differences between the conditions in (a), (c), and (g) and those in (b), (e), and (/)?
5. Show that {(1, 1,0, 0), (1, 0, 1, 1)} and {(2, 1, 3, 3), (0, 1, 1, 1)} span
the same subspace of R 4 .
6. Let W be the subspace of R 5 spanned by {(1,1,1,1,1), (1,0,1,0,1),
(0,1,1,1,0), (2,0,0,1,1), (2,1,1,2,1), (1, 1, 1, 2,2), (1,2,3,4, 1)}.
Find a basis for W and the dimension of W.
7. Show that {(1, 1,2, 3), (1, 1, 2, 0), (3, 1,6, 6)} and {(1,0,1,0),
(0, 2, 0, 3)} do not span the same subspace.
8. Let W = <(l, 2, 3, 6), (4, 1,3, 6), (5, 1, 6, 12)> and W 2 = <(1, 1, 1, 1),
(2, 1,4, 5)> be subspaces of R 4 . Find bases for W x n W 2 and W x + W 2 . Extend
the basis of W x n W 2 to a basis of W lt and extend the basis of W x n VV 2 to a basis
of W 2 . From these bases obtain a basis of W x + W 2 .
9. Let P be the space of all polynomials with real coefficients, and let W ± =
(p(x) \p(l) = 0} and W 2 = { p (x) \p(l) = 0}. Determine W 1 n W 2 and W x + W 2 .
(These spaces are infinite dimensional and the student is not expected to find
bases for these subspaces. What is expected is a simple criterion or description
of these subspaces.)
10. We have already seen (Section 1, Exercise 11) that the real numbers form
a vector space over the rationals. Show that {1, V2} and {1  V2, 1 + V2}
span the same subspace.
11. Show that if W x and W 2 are subspaces, then W x u W 2 is not a subspace
unless one is a subspace of the other.
12. Show that the set of all vectors (x lt x 2 , x 3 , x 4 ) e R 4 satisfying the equations
3x x — 2x 2 — x 3 — 4x 4 =
is a subspace of R 4 . Find a basis for this subspace. {Hint: Solve the equations for
x x and x 2 in terms of x s and a; 4 . Then specify various values for x z and x 4 to obtain
as many linearly independent vectors as are needed.)
13. Let S, T, and T* be three subspaces of V (of finite dimension) for which
(a)S n T = S nT*,(b)S + T = S +T*, (c) T c T*. Show that T = T*.
14. Show by example that it is possible to have S ®T = S ®T* without having
T = T*.
15. If V = Wj e W 2 and W is any subspace of V such that W x c W, show that
W = (W nWj) (W n VV 2 ). Show by an example that the condition W x <= W
(or W 2 <= VV) is necessary.
chapter
II
Linear
transformations
and matrices
In this chapter we define linear transformations and various operations:
addition of two linear transformations, multiplication of two linear trans
formations, and multiplication of a linear transformation by a scalar.
Linear transformations are functions of vectors in one vector space U
with values which are vectors in the same or another vector space V which
preserve linear combinations. They can be represented by matrices in the
same sense that vectors can be represented by wtuples. This representation
requires that operations of addition, multiplication, and scalar multiplication
of matrices be defined to correspond to these operations with linear trans
formations. Thus we establish an algebra of matrices by means of the
conceptually simpler algebra of linear transformations.
The matrix representing a linear transformation of U into V depends on
the choice of a basis in U and a basis in V. Our first problem, a recurrent
problem whenever matrices are used to represent anything, is to see how a
change in the choice of bases determines a corresponding change in the
matrix representing the linear transformation. Two matrices which represent
the same linear transformation with respect to different sets of bases must
have some properties in common. This leads to the idea of equivalence
relations among matrices. The exact nature of this equivalence relation
depends on the bases which are permitted.
In this chapter no restriction is placed on the bases which are permitted
and we obtain the widest kind of equivalence. In Chapter III we identify
U and V and require that the same basis be used in both. This yields a
more restricted kind of equivalence, and a study of this equivalence is both
interesting and fruitful. In Chapter V we make further restrictions in the
permissible bases and obtain an even more restricted equivalence.
When no restriction is placed on the bases which are permitted, the
26
1 I Linear Transformations 27
equivalence is so broad that it is relatively uninteresting. Very useful results
are obtained, however, when we are permitted to change basis only in the
image space V. In every set of mutually equivalent matrices we select one,
representative of all of them, which we call a normal form, in this case
the Hermite normal form. The Hermite normal form is one of our most
important and effective computational tools, far exceeding in utility its
application to the study of this particular equivalence relation.
The pattern we have described is worth conscious notice since it is re
current and the principal underlying theme in this exposition of matrix
theory. We define a concept, find a representation suitable for effective
computation, change bases to see how this change affects the representation,
and then seek a normal form in each class of equivalent representations.
1 I Linear Transformations
Let U and V be vector spaces over the same field of scalars F.
Definition. A linear transformation a of U into V is a singlevalued mapping
of U into V which associates to each element a e U a unique element <r(a) e V
such that for all a, e U and all a, b e F we have
o(a<x + bp) = aa(<x) + bo(P). (1.1)
We call cr(a) the image of a under the linear transformation a. If a e V,
then any vector <xeU such that <r(a) = a is called an inverse image of a.
The set of all a e U such that <r(<x) = a is called the complete inverse image
of a, and it is denoted by o r_1 (a). Generally, <r 1 (a) need not be a single
element as there may be more than one aeli such that c(a) = a.
By taking particular choices for a and b we see that for a linear trans
formation cr(a + 8) = ff(a) + <r(/8) and (r(aa) = a(r(a). Loosely speaking,
the image of the sum is the sum of the images and the image of the product
is the product of the images. This descriptive language has to be interpreted
generously since the operations before and after applying the linear trans
formation may take place in different vector spaces. Furthermore, the remark
about scalar multiplication is inexact since we do not apply the linear trans
formation to scalars; the linear transformation is defined only for vectors
in U. Even so, the linear transformation does preserve the structural
operations in a vector space and this is the reason for its importance. Gener
ally, in algebra a structurepreserving mapping is called a homomorphism.
To describe the special role of the elements of F in the condition, a(aa.) =
ao(<x), we say that a linear transformation is a homomorphism over F, or an
Fhomomorphism .
If for a 5^ j8 it necessarily follows that <r(a) ^ o(ji), the homomorphism a
is said to be onetoone and it is called a monomorphism. If A is any subset of
28 Linear Transformations and Matrices  II
U, a(A) will denote the set of all images of elements of A; a{A) = {a  a =
<r(a) for some a e A}. o(A) is called the image of A. a(U) is often denoted by
Im((r) and is called the image of a. If Im(or) = V we shall say that the homo
morphism is a mapping onto V and it is called an epimorphism.
We call the set U, on which the linear transformation a is defined, the
domain of a. We call V, the set in which the images of a are defined, the
codomain of a. Strictly speaking, a linear transformation must specify
the domain and codomain as well as the mapping. For example, consider
the linear transformation that maps every vector of U onto the zero vector of
V. This mapping is called the zero mapping. If W is any subspace of V, there is
also a zero mapping of U into W, and this mapping has the same effect on the
elements of U as the zero mapping of U into V. However, they are different
linear transformations since they have different codomains. This may seem
like an unnecessarily fine distinction. Actually, for most of this book we
could get along without this degree of precision. But the more deeply we go
into linear algebra the more such precision is needed. In this book we need
this much care when we discuss dual spaces and dual transformations in
Chapter IV.
A homomorphism that is both an epimorphism and a monomorphism is
called an isomorphism. If a e V, the fact that a is an epimorphism says that
There is an a e U such that a(<x) = a. The fact that a is a monomorphism says
that this a is unique. Thus, for an isomorphism, we can define an inverse
mapping cr _1 that maps a onto a.
Theorem 1.1. The inverse cr 1 of an isomorphism is also an isomorphism.
proof. Since a' 1 is obviously onetoone and onto, it is necessary only
to show that it is linear. If a = <r(a) and /?= <r(/3), then a(aa. + bp) =
a* + bp so that a\adL + bfi) = a<x + bp = acr\a) + ba 1 ^). □
For the inverse isomorphism cr _1 (a) is an element of U. This conflicts with
the previously given definition of ff _1 (a) as a complete inverse image in which
o\dL) is a subset of U. However, the symbol cr 1 , standing alone, will
always be used to denote an isomorphism, and in this case there is no diffi
culty caused by the fact that <r _1 (a) might denote either an element or a one
element set,
Let us give some examples of linear transformations. Let U = V = P,
the space of polynomials in x with coefficients in R. For a = 27=0°^'
define cr(a) = — = 2 n =o "**?■ " ln calculus one of tne very first things
dx l % ' d(* + 0) da dp J
proved about the derivative is that it is linear, = — + — and
j/ \ j dx dx dx
— — = a — . The mapping t(<x) = ^Lo  — ^7 x * +1 is also linear  Notice
dx dx i + 1
that this is not the indefinite integral since we have specified that the constant
1 I Linear Transformations 29
of integration shall be zero. Notice that a is onto but not onetoone and t
is onetoone but not onto.
Let U = R n and V = R m with m < n. For each a = (a lt . . . , a n ) e R n
define a(a) = (a l5 . . . , a TO ) e R m . It is clear that this linear transformation
is onetoone if and only if m = n, but it is onto. For each /? = {b x , . . . ,
bj g R m define t(j8) = fo, . . . , 6 m , 0, . . . , 0) e R n . This linear transforma
tion is onetoone, but it is onto if and only if m = n.
Let U = V. For a given scalar ae F the mapping of a onto a a is linear since
fl(a + 0) = aa + a£ = a(a) + a(0),
and
«(&<x) = (a6)a = (ba)<x. = b ■ a(cn).
To simplify notation we also denote this lineartransformation by a. Linear
transformations of this type are called scalar transformations, and there is a
onetoone correspondence between the field of scalars and the set of scalar
transformations. In particular, the linear transformation that leaves every
vector fixed is denoted by 1. It is called the identity transformation or unit
transformation. If linear transformations in several vector spaces are being
discussed at the same time, it may be desirable to identify the space on which
the identity transformation is defined. Thus ly will denote the identity
transformation on U.
When a basis of a finite dimensional vector space V is used to establish a
correspondence between vectors in V and «tuples in F n , this correspondence
is an isomorphism. The required arguments have already been given in
Section 13. Since V and F n are isomorphic, it is theoretically possible to
discuss the properties of V by examining the properties of F n . However, there
is much interest and importance attached to concepts that are independent
of the choice of a basis. If a homomorphism or isomorphism can be defined
uniquely by intrinsic properties independent of a choice of basis the mapping is
said to be natural or canonical. In particular, any two vector spaces of
dimension n over F are isomorphic. Such an isomorphism can be established
by setting up an isomorphism between each one and F n . This isomorphism
will be dependent on a choice of a basis in each space. Such an isomorphism,
dependent upon the arbitrary choice of bases, is not canonical.
Next, let us define the various operations between linear transformations.
For each pair a, r of linear transformation of U into V, define a + t by the
rule
(a + t)(<x) = <r(a) + t(oc) for all ae^
a + t is a linear transformation since
(a + r)(aa + bfi) = o(a«. + bfi) + r(aa + bp) = aa{a) + M0)
+ aT(a) + M0) = a[o(a) + r(a)] + b{a{fi) + r(/3)]
= a(a + t)(oc) + b(a + t)(jS).
30 Linear Transformations and Matrices  II
Observe that addition of linear transformation is commutative ; a + r =
r + a.
For each linear transformation a and a e F define aa by the rule ; (aa)(ct) =
a[a(<x.)]. aa is a linear transformation.
It is not difficult to show that with these two operations the set of all linear
transformations of U into V is itself a vector space over F. This is a very
important fact and we occasionally refer to it and make use of it. However,
we wish to emphasize that we define the sum of two linear transformations
if and only if they both have the same domain and the same codomain.
It is neither necessary nor sufficient that they have the same image, or that the
image of one be a subset of the image of the other. It is simply a question of
being clear about the terminology and its meaning. The set of all linear
transformations of U into V will be denoted by Hom(U, V).
There is another, entirely new, operation that we need to define. Let W
be a third vector space over F. Let a be a linear transformation of U into V
and t a linear transformation of V into W. By ra we denote the linear trans
formation of U into W defined by the rule: (to)(oc) = r[a(oC)]. Notice that
in this context ar has no meaning. We refer to this operation as either
iteration or multiplication of linear transformation, and ra is called the
product of t and a.
The operations between linear transformations are related by the following
rules :
1. Multiplication is associative: tt{to) = (rrr)a. Here n is a linear
transformation of W into a fourth vector space X.
2. Multiplication is distributive with respect to addition:
(ti + t 2 )o = r x a + r 2 a and r(a 1 + a 2 ) = ra x + tg 2 .
3. Scalar multiplication commutes with multiplication: a(TG) = r(aa).
These properties are easily proved and are left to the reader.
Notice that if W # U, then to is defined but or is not. If all linear trans
formations under consideration are mappings of a vector space U into itself,
then these linear transformations can be multiplied in any order. This means
that ra and ar would both be defined, but it would not mean that ra = ar.
The set of linear transformation of a vector space into itself is a vector
space, as we have already observed, and now we have defined a product
which satisfies the three conditions given above. Such a space is called an
associative algebra. In our case the algebra consists of linear transformation
and it is known as a linear algebra. However, the use of terms is always in
a state of flux, and today this term is used in a more inclusive sense. When
referring to a particular set with an algebraic structure, "linear algebra"
still denotes what we have just described. But when referring to an area of
1 I Linear Transformations 31
study, the term "linear algebra includes virtually every concept in which
linear transformations play a role, including linear transformations between
different vector spaces (in which the linear transformations cannot always
be multiplied), sequences of vector spaces, and even mappings of sets of
linear transformations (since they also have the structure of a vector space).
Theorem 1.2. Im(o) is a subspace of V.
proof. If a and j5 are elements of Im(cr), there exist a, e U such that
<r(a) = a and c(p) = /5. For any a, beF, a(a<x + b§) = aa(<x.) + ba(fi) =
av. + bfi g Im(cr). Thus Im(cr) is a subspace of V. D
Corollary 1.3. If L/ x is a subspace of U, then (ji^d is a subspace of V. □
It follows from this corollary that <r(0) = where denotes the zero vector
of U and the zero vector of V. It is even easier, however, to show it directly.
Since o"(0) = <r(0 + 0) = tf(0) + <r(0) it follows from the uniqueness of the
zero vector that cr(0) = 0.
For the rest of this book, unless specific comment is made, we assume
that all vector spaces under consideration are finite dimensional. Let
dim U = n and dim V = m.
The dimension of the subspace Im(cr) is called the rank of the linear trans
formation a. The rank of a is denoted by p(ff).
Theorem 1.4. p($ < {m, n}.
proof. If {<*!, . . . , <x s } is linearly dependent in U, there exists a nontrivial
relation of the form 2< a^ = 0. But then 2i «i^( a ») = ff (°) = 0; that is,
{ofay), . . . , cr(a s )} is linearly dependent in V. A linear transformation
preserves lin ear relations and transforms dependent sets into dependent
sets. Thus, there can be no more than n linearly independent elements in
Im((r). In addition, Im(<r) is a subspace of V so that dim lm(a) < m. Thus
p(o) = dim Im(o) < min {m, n). □
Theorem 1.5. If W is a subspace of V, the set <r\W) of all a e U such that
(r(a) £ W is a subspace of U.
proof. If a, 6 (f^W), then a(ax + bfS) = ao{a) + botf) e W. Thus
au. + bp e <r 1 (W) and tr^W) is a subspace. □
The subspace K(a) = a _1 (0) is called the kernel of the linear transformation
a. The dimension of K{a) is called the nullity of a. The nullity of a is denoted
by v(a).
Theorem 1.6. p(<r) + v(<r) = n.
proof. Let {a l5 . . . , a v , &, . . . , p k ) be a basis of U such that {a l9 . . . , a v }
is a basis of K(a). For a = 2* a^ + 2, bfi s e U we see that <r(a) =
2, a*tf(oO + L MA) = 2* W Thus Wi), ■••, *(&)} spans Im(cr).
On the other hand if % c,<r(ft) = 0, then cr(2, c,&) = 2, c/Kft) = 0; that
32 Linear Transformations and Matrices  II
is, 2* c Si e K ( a ) In tnis case tnere exist coefficients d { such that 2; c>& =
2, di&i If any of these coefficients were nonzero we would have a non
trivial relation among the elements of {a l5 . . . , a,, ^ l5 . . . , /?*.}. Hence, all
c f = and M/^), . . . , cr(0 fc )} is linearly independent. But then it is a basis
of Im(<r) so that k = p(a). Thus p{a) + r(a) = n. U
Theorem 1.6 has an important geometric interpretation. Suppose that
a 3dimensional vector space R 3 were mapped onto a 2dimensional vector
space R 2 . In this case, it is simplest and sufficiently accurate to think of a
as the linear transformation which maps (a u a 2 , a 3 ) e R 3 onto (a lt a 2 ) e R 2
which we can identify with (a lt a 2 , 0) e R 3 . Since p(a) = 2, i>(V) = 1.
Clearly, every point (0, 0, « 3 ) on the :r 3 axis is mapped onto the origin.
Thus K{a) is the z 3 axis, the line through the origin in the direction of the
projection, and {(0, 0, 1) = aj is a basis of K(a). It should be evident
that any plane through the origin not containing K{o) will be projected onto
the a^plane and that this mapping is onetoone and onto. Thus the com
plementary subspace <&, /S 2 > can be taken to be any plane through the origin
not containing the rr 3 axis. This illustrates the wide latitude of choice possible
for the complementary subspace </? l5 . . . , P p ).
Theorem 1.7. A linear transformation a of U into V is a monomorphism if
and only ifv{a) = 0, and it is an epimorphism if and only if p(a) = dim V.
proof. K(a) = {0} if and only if v(o) = 0. If a is a monomorphism, then
certainly K{o) = {0} and v{a) = 0. On the other hand, if v(a) = and
<r(a) = ff(j8), then <r(a  j8) = so that a  e K(o) = {0}. Thus, if
v(a) = 0, a is a monomorphism.
It is but a matter of reading the definitions to see that o is an epimorphism
if and only if p(a) = dim V. □
If dim U = n < dim V = m, then p(a) = n — v(a) < n < m so that
a cannot be an epimorphism. If« > m,thenv(a) = n — p(a) >n — m> 0,
so that a cannot be a monomorphism. Any linear transformation from a
vector space into a vector space of higher dimension must fail to be an
epimorphism. Any linear transformation from a vector space into a vector
space of lower dimension must fail to be a monomorphism.
Theorem 1.8. Let U and V have the same finite dimension n. A linear
transformation a ofil into V is an isomorphism if and only if it is an epimorphism.
a is an isomorphism if and only if it is a monomorphism.
proof. It is part of the definition of an isomorphism that it is both an
epimorphism and a monomorphism. Suppose a is an epimorphism. p(a) =
n and v(o) = by Theorem 1.6. Hence, a is a monomorphism. Conversely
if a is a monomorphism, then v(a) = and, by Theorem 1.6, p(o) = n.
Hence, a is an epimorphism. □
1 I Linear Transformations 33
Thus a linear transformation a of U into V is an isomorphism if two of the
following three conditions are satisfied: (1) dim U = dim V, (2) a is an
epimorphism, (3) a is a monomorphism.
Theorem 1.9. p(r) = p(ro) + dim (Im(<7) n A"(r)}.
proof. Let t' be a new linear transformation defined on lm(a) mapping
ImO) into W so that for all a e Im(o), t'(oc) = T (a). Then AT(t') = Im(or) n
^(t) and />(Y) = dim t [Im(»] = dim r<r(L/) = p{ra). Then Theorem 1.6
takes the form
p(r') + v(t') *= dim Im(o),
or
p(r<r) + dim (Im(<r) n #(t)} = p{a). D
Corollary 1.10. p(ra) = dim (Im(o) + K(r)}  v(t).
proof. This follows from Theorem 1.9 by application of Theorem 4.8
of Chapter I. □
Corollary 1.11. IfK(r) <= Im(cr), then p(a) = p(ra) + v(r). □
Theorem 1.12. The rank of a product of linear transformations is less than
or equal to the rank of either factor : p{ra) < min {p(r), p{a)}.
proof. The rank of to is the dimension of r[a(U)] c T (V). Thus consider
ing dim <r(U) as the ' V and dim t(V) as the "m" of Theorem 1.3 we see that
dim ra(U) = p{ra) < min {dim or(V), dim t (V)} = min {p{a), p(r)}. D
Theorem 1.13. If a is an epimorphism, then p(ja) = p(r). If r is a mono
morphism, then p{ra) = p(a).
proof. If a is an epimorphism, then K(t) <= Im(o) = V and Corollary
1.11 applies. Thus p(ra) = p{a) — v(t) = m — v(t) = p(r). If t is a
monomorphism, then K(r) = {0} <= Im(o) and Corollary 1.11 applies.
Thus p(ra) = p(a)  v(t) = p(a). D
Corollary 1.14. The rank of a linear transformation is not changed by
multiplication by an isomorphism {on either side). □
Theorem 1.15. a is an epimorphism if and only if to = implies r — 0.
t i5 a monomorphism if and only ifra = implies a = 0.
proof. Suppose a is an epimorphism. Assume to is defined and ra = 0.
If t 5^ 0, there is a e V such that t(8) 5^ 0. Since cr is an epimorphism,
there is an a e U such that <r(a) = 0. Then T<r(a) = t(^) 5^ 0. This is a
contradiction and hence t = 0. Now, suppose tot = implies t = 0. If or is
not an epimorphism then Im(cr) is a subspace of V but Im(a) ^ V. Let
{/?!, . . . , &.} be a basis of Im(o), and extend this independent set to a basis
{ft, . . . , 0„ . . . , PJ of V. Define r^) = ft for » > r and r(ft) = for
34 Linear Transformations and Matrices  II
/ <, r. Then to = and t^O. This is a contradiction and, hence, a is an
epimorphism.
Now, assume ra is defined and ra = 0. Suppose r is a monomorphism.
If a 5^ 0, there is an a e U such that <r(a) # 0. Since t is a monomorphism,
T<r(a) 5^ 0. This is a contradiction and, hence, a = 0. Now assume rer =
implies <r = 0. If r is not a monomorphism there is an a e L/ such that a^0
and t(<x) = 0. Let {vl x , . . . , a„} be any basis of U. Define ^(a,) = a for each
i. Then To'(a l ) = r(a) = for all / and ra = 0. This is a contradiction and,
hence, r is a monomorphism. □
Corollary 1.16. a is an epimorphism if and only if r x a = r 2 a implies
T i = T 2 T w a monomorphism if and only if ra x = ra 2 implies a x = a 2 .
The statement that r x a = r 2 a implies t x = r 2 is called a rightcancellation,
and the statement that ra x = ro 2 implies a x = cr 2 is called a leftcancellation.
Thus, an epimorphism is a linear transformation that can be cancelled on the
right, and a monomorphism is a linear transformation that can be cancelled
on the left.
Theorem 1.17. Let A = {<x l5 . . . , a J be any basis of U. Let 6 = {/5j, . . . ,
/3J be any n vectors in V (not necessarily linearly independent). There exists
a uniquely determined linear transformation a of U into V such that o^a,) = /9 t
for i= 1,2, ... ,n.
proof. Since A is a basis of U, any vector ccg U can be expressed uniquely
in the form a = ^" =1 <*&■ If <* is to be linear we must have
n n
■■'£ *,** ;! (7(a) = ^ *<<<«)< = 2 «ift e U.
It is a simple matter to verify that the mapping so defined is linear. □
Corollary 1.18. Let C = {y lt . . . , y r ) be any linearly independent set in U,
where U is finite dimensional. Let D — {d x , . . . , d r ) be any r vectors in V.
There exists a linear transformation a of U into V such that ^(y^) = d t for
i = 1 , . . . , r.
proof. Extend C to a basis of U. Define a(y t ) = d t for i = 1, . . . , r,
and define the values of a on the other elements of the basis arbitrarily. This
will yield a linear transformation a with the desired properties. □
It should be clear that, if C is not already a basis, there are many ways to
define a. It is worth pointing out that the independence of the set C is crucial
to proving the existence of the linear transformation with the desired prop
erties. Otherwise, a linear relation among the elements of C would impose
a corresponding linear relation among the elements of D, which would mean
that D could not be arbitrary.
1 I Linear Transformations 35
Theorem 1.17 establishes, for one thing, that linear transformations really
do exist. Moreover, they exist in abundance. The real utility of this theorem
and its corollary is that it enables us to establish the existence of a linear
transformation with some desirable property with great convenience. All
we have to do is to define this function on an independent set.
Definition. A linear transformation tt of V into itself with the property that
7T 2 = 7T is called a. projection.
Theorem 1.19. If tt is a projection oj " V into itself, then V = lm(rr) © K(tt)
and rr acts like the identity on Im(7r).
proof. For a G V, let o^ = 7r(a). Then 7r(ai) = 7r 2 (a) = 7r(a) = <*j. This
shows that tt acts like the identity on IrmV). Let <x 2 = a — a x . Then 7r(a 2 ) =
7r(a) — Trfai) = a x — a x = 0. Thus a = a x + ol 2 where x 1 e Im(7r) and
<x 2 g K(tt). Clearly, Im(7r) n K(tt) = {0}. □
Fig. 1
If S =' Im(7r) and T = K(tt), we say that tt is a projection of V onto S along
T. In the case where V is the real plane, Fig. 1 indicates the interpretation of
these words, a is projected onto a point of S in a direction parallel to 7*.
EXERCISES
1. Show that o((x 1 ,x 2 )) = (x 2 , x x ) defines a linear transformation of R 2 into
itself.
2. Let #!((#!, # 2 )) = 0e 2 , — x x ) and ^((^i* ^2)) = (^i> ~ x 2) Determine a x + a 2 ,
G\G 2 and (TgfTj.
3. Let U = V = R™ and let a^, a; 2 » • • • » x »)) = (^1. « 2 > • • • > x fc> 0, • • • , 0)
where k < n. Describe Im(cr) and A^(ct).
4. Let o((x lt x 2> %> %)) = Q x i — 2cc 2 — x 3 — 4x 4 , x x + x 2 — 2x 3 — 3%). Show
that a is a linear transformation. Determine the kernel of a.
5. Let o((x lt x 2 , x 3 )) = (2^ + x 2 + 3x 3 , 3x x — x 2 + x 3 , —\x x + 3x 2 + x 3 ). Find
36 Linear Transformations and Matrices  II
a basis of o(U). (Hint: Take particular values of the x t to find a spanning set for
or(U).) Find a basis of K(o).
6. Let D denote the operator of differentiation,
dy d 2 v
D(y) = j x , D\y) = D[D(y)] = ^ , etc.
Show that D n is a linear transformation, and also that p(D) is a linear transforma
tion if p(D) is a polynomial in D with constant coefficients. (Here we must assume
that the space of functions on which D is denned contains only functions differen
tiable at least as often as the degree of p(D).)
7. Let U = V and let o and t be linear transformations of U into itself. In this
case gt and to are both defined. Construct an example to show that it is not
always true that or = to.
8. Let U = V = P, the space of polynomials in x with coefficients in R. For
" = 2,7=0 ^ let
n
tfC"*) = 2 * a i xil
and
i=0 * "I" 1
Show that <tt = 1, but that to ^ 1.
9. Show that if two scalar transformations coincide on U then the defining
scalars are equal.
10. Let o be a linear transformation of U into V and let A = {a x , . . . , a„} be a
basis of U. Show that if the values {0(0.^}, . . . , o(a. n )} are known, then the value
of <r(a) can be computed for each a e U.
11. Let 1/ and V be vector spaces of dimensions n and w, respectively, over the
same field F. We have already commented that the set of all linear transformations
of U into V forms a vector space. Give the details of the proof of this assertion.
Let A = {04, . . . , a n } be a basis of U and 8 = {/? x , . . . , /? TO } be a basis of V. Let o^
be the linear transformation of U into V such that
[0 if k *j,
[& if k =j.
Show that {tr j; I / — 1, . . . , m;j = 1, . . . , n} is a basis of this vector space.
For the following sequence of problems let dim U = n and dim V — m. Let o
be a linear transformation of U into V and t a linear transformation of V into W.
12. Show that p(a) < P ( r o) + i>(t). (///«/: Let V = o(U) and apply Theorem
1.6 to t defined on V.)
13. Show that max {0, p(o) + p(r) — m} < p(to) < min (p(t), p(<r)}.
2 I Matrices
37
14. Show that max {n — m + v(t), v(o)} <, v(to) <, min {n, v(o) + v( T )}. (For
m = n this inequality is known as Sylvester's law of nullity.)
15. Show that if j>(t) = 0, then p (to) = P (a).
16. It is not generally true that v(a) = implies P (ra) = P ( T ). Construct an
example to illustrate this fact. (Hint: Let m be very large.)
17. Show that if m = n and v(a) = 0, then p (to) = p (t).
18. Show that if a x and <r 2 are linear transformations of U into V, then
P (a 1 + a 2 ) < min {m, n, pK) + P (a z )}.
19. Show that />Oi)  p(ff 2 )l ^ p(°i + ff 2)
20. If S is any subspace of V there is a subspace T such that V = S © T. Then
every a; e V can be represented uniquely in the form a = a x + a 2 where a x e S and
a 2 G T. Show that the mapping rr which maps a onto a x is a linear transformation.
Show that T is the kernel of v. Show that tt 2 = w. The mapping tt is called a
projection of V onto S along 7.
21. (Continuation) Let n be a projection. Show that 1 — tt is also a projection.
What is the kernel of 1 — nl Onto what subspace is 1 — tt a projection? Show
that n(l  n) =0.
2 I Matrices
Definition. A matrix over a field F is a rectangular array of scalars. The
array will be written in the form
02i a 9
(2.1)
whenever we wish to display all the elements in the array or show the form
of the array. A matrix with m rows and n columns is called an m x n
matrix. Ann x n matrix is said to be of order n.
We often abbreviate a matrix written in the form above to [a ti ] where
the first index denotes the number of the row and the second index denotes
the number of the column. The particular letter appearing in each index
position is immaterial; it is the position that is important. With this con
vention a H is a scalar and [a i} ] is a matrix. Whereas the elements a H and a kl
need not be equal, we consider the matrices [a i} ] and [a kl ] to be identical
since both [a i} ] and [a kl ] stand for the entire matrix. As a further convenience
we often use upper case Latin italic letters to denote matrices; A = [a u ].
Whenever we use lower case Latin italic letters to denote the scalars appearing
38 Linear Transformations and Matrices  II
in the matrix, we use the corresponding upper case Latin italic letter to denote
the matrix. The matrix in which all scalars are zero is denoted by (the third
use of this symbol!). The a tj appearing in the array [a i} \ are called the
elements of [a ti ]. Two matrices are equal if and only if they have exactly
the same elements. The main diagonal of the matrix [a {j ] is the set of elements
{a n , . . . , a n ) where / = min {m, n}. A diagonal matrix is a square matrix
in which the elements not in the main diagonal are zero.
Matrices can be used to represent a variety of different mathematical con
cepts. The way matrices are manipulated depends on the objects which they
represent. Considering the wide variety of situations in which matrices have
found application, there is a remarkable similarity in the operations performed
on matrices in these situations. There are differences too, however, and to
understand these differences we must understand the object represented and
what information can be expected by manipulating with the matrices. We
first investigate the properties of matrices as representations of linear trans
formations. Not only do the matrices provide us with a convenient means of
doing whatever computation is necessary with linear transformations, but the
theory of vector spaces and linear transformations also proves to be a power
ful tool in developing the properties of matrices.
Let U be a vector space of dimension n and V a vector space of dimension
m, both over the same field F. Let A = {a x , . . . , a n } be an arbitrary but
fixed basis of U, and let B = {j8 l5 . . . , /S TO } be an arbitrary but fixed basis
of V. Let a be a linear transformation of U into V. Since c(a,) e V, <r(a,)
can be expressed uniquely as a linear combination of the elements of B ;
m
<*«,) = I <»<A (22)
We define the matrix representing a with respect to the bases A and B to be
the matrix A = [a^].
The correspondence between linear transformations and matrices is
actually onetoone and onto. Given the linear transformation cr, the a tj
exist because B spans V, and they are unique because B is linearly independent.
On the other hand, let A = [a i} ] be any m x n matrix. We can define
o(ol } ) = 2*Li a aPi f° r each oijE A, and then we can extend the proposed
linear transformation to all of U by the condition that it be linear. Thus,
if I = 2j=i x ; a i' we define #« &***« ^uJ.;»4u'i«  • •*. "!.,*•.<*. /.'•'"■'
n i m \
m I n \
= 2(2««*i)A. (23)
i=i y=i /
2 I Matrices
39
a can be extended to all of U because A spans U, and the result is well defined
(unique) because A is linearly independent.
Here are some examples of linear transformations and the matrices which
represent them. Consider the real plane R 2 = U = V. Let A = 6 = {(1,0),
(0, 1)}. A 90° rotation counterclockwise would send (1,0) onto (0, 1) and
it would send (0, 1) onto (1,0). Since <r((l, 0)) = • (1, 0) + 1 • (0, 1) and
(T((0, 1)) = (1) • (1, 0) + • (0, 1), cr is represented by the matrix
1
1
The elements appearing in a column are the coordinates of each image of a
basis vector under a transformation.
In general, a rotation counterclockwise through an angle of 6 will send
(1, 0) onto (cos 6, sin 0) and (0, 1) onto (—sin 0, cos 0). Thus this rotation
is represented by
cos
sin (
sin
cos
(2.4)
Suppose now that r is another linear transformation of U into V represented
by the matrix B = [b ti ]. Then for the sum a + t we have
(a + r)(a,) = cr(a,) + r(a,) = J a w ft + J b it fi t
m
= 2 ( a ii + b ii)Pi
(2.5)
Thus a + t is represented by the matrix [a {j + Z>„]. Accordingly, we
define the sum of two matrices to be that matrix obtained by the addition
of the corresponding elements in the two arrays; A + B = [a tj + b u \ is
the matrix corresponding to a + t. The sum of two matrices is denned if
and only if the two matrices have the same number of rows and the same
number of columns.
If a is any scalar, for the linear transformation aa we have
(0<r)(a,.) = a 2 a„ft = 2 (™a)P<
(2.6)
Thus aa is represented by the matrix [aa^]. We therefore define scalar
multiplication by the rule a A = [aa^].
Let W be a third vector space of dimension r over the field F, and let
C = {yi» • • • , y r } be an arbitrary but fixed basis of W. If the linear trans
formation a of U into V is represented by them xn matrix A = [a^] and the
40
Linear Transformations and Matrices I II
linear transformation t of V into W is represented by the r x m matrix
B = [b ki ], what matrix represents the linear transformation to of U into W?
(V(7)(a,) = t{o{<x,)) = rl >u&)
m
[_0^v — .
ecin,
V$jp*4UV*>~h^ <A &? v
7H j T \
r /m \
fc=l \i=l //
S.a.^,1
• (2.7)
u.»vAu I p 1
Thus, if we define c kj = ^T =1 b^a^, then C = [c kj ] is the matrix representing
the product transformation to. Accordingly, we call C the matrix product
of B and >4, in that order: C = BA. ^J u^C**** ^.k*>.i Ike ^«jt <
For computational purposes it is customary to write the arrays of B ^T" 1
and A side by side. The element c kj of the product is then obtained by ,^ 7<
multiplying the corresponding elements of row k of B and column j of A
and adding. We can trace the elements of row k of B with a finger of the
left hand while at the same time tracing the elements of column j of A with
a finger of the right hand. At each step we compute the product of the
corresponding elements and accumulate the sum as we go along. Using
this simple rule we can, with practice, become quite proficient, even to the
point of doing "without hands."
Check the process in the following examples :
1 4
1
2
1
2 1
2
"l
f
2"
2
" 5
2"
3
2
1
=
11
1
2_
2_
3
2
All definitions and properties we have established for linear transformations
can be carried over immediately for matrices. For example, we have:
1. • A = 0. (The "0" on the left is a scalar, the "0" on the right is a
matrix with the same number of rows and columns as A.)
2. \ A = A.
3. A(B + C)= AB + AC.
4. {A + B)C = AC + BC.
5. A(BC) = (AB)C.
Of course, in each of the above statements we must assume the operations
proposed are well defined. For example, in 3, B and C must be the same
2 I Matrices
41
size and A must have the same number of columns as B and C have rows.
The rank and nullity of a matrix A are the rank and nullity of the associated
linear transformation, respectively.
Theorem 2.1. For an m x n matrix A, the rank of A plus the nullity of A
is equal to n. The rank of a product BA is less than or equal to the rank of
either factor.
These statements have been established for linear transformations and
therefore hold for their corresponding matrices. □
The rank of a is the dimension of the subspace Im(<r) of V. Since Im(a)
is spanned by {ofaj, . . . , <r(a B )}, p{a) is the number of elements in a
maximal linearly independent subset of {</(%), . . . , <r(a w )}. Expressed
in terms of coordinates, <r(a,) = J™ =1 a it p t is represented by the mtuple
(a li5 a 2j , . . . , a mj ), which is the ratuple in column j of the matrix [a t ].
Thus p{a) = p{A) is also equal to the maximum number of linearly inde
pendent columns of A. This is usually called the column rank of a matrix
A, and the maximum number of linearly independent rows of A is called
the row rank of A. We, however, show before long that the number of
linearly independent rows in a matrix is equal to the number of linearly
independent columns. Until that time we consider "rank" and "column
rank" as synonymous.
Returning to Equation (2.3), we see that, if £ e 1/ is represented by
( x i, ■ • • , * n ) and the linear transformation a of U into V is represented
by the matrix A = [a ti ], then <r(£) e V is represented by (y ls . . . , yj where
yt=i a n x i (i=l, ...,m).
3=1
(2.8)
In view of the definition of matrix multiplication given by Equation (2.7)
we can interpret Equations (2.8) as a matrix product of the form
where
r =
Vi
Y= AX
and
(2.9)
X =
This single matric equation contains the m equations in (2.8).
We have already used the ntuple (x lt . . . , x n ) to represent the vector
f = XU x i<*i Because of the usefulness of equation (2.9) we also find it
convenient to represent f by the onecolumn matrix X. In fact, since it is
42
Linear Transformations and Matrices I II
somewhat wasteful of space and otherwise awkward to display onecolumn
matrices we use the «tuple (x 1} . . . , x n ) to represent not only the vector £
but also the column matrix X. With this convention [x x • • • x n ] is a onerow
matrix and (x lt . . . , x n ) is a onecolumn matrix.
Notice that we have now used matrices for two different purposes, (1) to
represent linear transformations, and (2) to represent vectors. The single
matric equation Y = AX contains some matrices used in each way.
EXERCISES
1 . Verify the matrix multiplication in the following examples :
(«)
(*)
3
5
1 2"
2 3
' 2
1
1
' 2
1
1
1
3"
6
1
=
2
3
9
1
3"
" 2"
"10"
6
1
3
=
15
2
1
4
(c)
2. Compute
3
5
3
9
4'
11
10"
15
4
2"
3
1
37
4
11
Interpret the answer to this problem in terms of the computations in Exercise 1.
3. Find AB and BA if
A =
"10 1"
110
10 10
10 1
B =
' 1
5
1
2 3
6 7
2 3
4
5 6 7 !
4. Let cr be a linear transformation of R 2 into itself that maps (1 , 0) onto (3, 1)
and (0, 1) onto ( — 1,2). Determine the matrix representing a with respect to the
bases A = B ={(1,0), (0,1)}.
2  Matrices 43
5. Let a be a linear transformation of R 2 into itself that maps (1, 1) onto (2, —3)
and (1, —1) onto (4, —7). Determine the matrix representing a with respect to the
bases A = B = {(1 , 0), (0, 1)}. (Hint: We must determine the effect of a when it is
applied to (1,0) and (0, 1). Use the fact that (1,0)= (1, 1) + 1(1, 1) and the
linearity of a.)
6. It happens that the linear transformation denned in Exercise 4 is onetoone,
that is, a does not map two different vectors onto the same vector. Thus, there is
a linear transformation that maps (3, 1) onto (1,0) and ( — 1,2) onto (0, 1).
This linear transformation reverses the mapping given by a. Determine the matrix
representing it with respect to the same bases.
7. Let us consider the geometric meaning of linear transformations. A linear
transformation of R 2 into itself leaves the origin fixed (why?) and maps straight
lines into straight lines. (The word "into" is required here because the image of a
straight line may be another straight line or it may be a single point.) Prove that
the image of a straight line is a subset of a straight line. (Hint: Let a be represented
by the matrix
A =
Then a maps (x, y) onto (a u a> + a 12 y, a 21 x + a 22 y). Now show that if (x, y)
satisfies the equation ax + by = c its image satisfies the equation
(aa 22  ba 21 )x + (a n b  a 12 a)y = (a u a 22 — a 12 a 21 )c.)
8. (Continuation) We say that a straight line is mapped onto itself if every
point on the line is mapped onto a point on the line (but not all onto the same
point) even though the points on the line may be moved around.
(a) A linear transformation maps (1, 0) onto (1,0) and (0, 1) onto (0, 1).
Show that every line through the origin is mapped onto itself. Show that each
such line is mapped onto itself with the sense of direction inverted. This linear
transformation is called an inversion with respect to the origin. Find the matrix
representing this linear transformation with respect to the basis {(1,0), (0, 1)}.
(b) A linear transformation maps (1,1) onto (1, 1) and leaves (1, 1)
fixed. Show that every line perpendicular to the line x x + x 2 = is mapped onto
itself with the sense of direction inverted. Show that every point on the line
x i + x 2 = is left fixed. Which lines through the origin are mapped onto them
selves? This linear transformation is called a reflection about the line x x + x 2 = 0.
Find the matrix representing this linear transformation with respect to the basis
{(1,0), (0, 1)}. Find the matrix representing this linear transformation with
respect to the basis {(1, 1), (1, —1)}.
(c) A liner transformation maps (1, 1) onto (2, 2) and (1, 1) onto (3, 3).
Show that the lines through the origin and passing through the points (1,1) and
(1, 1) are mapped onto themselves and that no other lines are mapped onto
themselves. Find the matrices representing this linear transformation with respect
to the bases {(1, 0), (0, 1)} and {(1, 1), (1, 1)}.
44
Linear Transformations and Matrices I II
(d) A linear transformation leaves (1 , 0) fixed and maps (0, 1) onto (1 , 1). Show
that each line x 2 = c is mapped onto itself and translated within itself a distance
equal to c. This linear transformation is called a shear. Which lines through the
origin are mapped onto themselves? Find the matrix representing this linear
transformation with respect to the basis {(1,0), (0, 1)}.
0) A linear transformation maps (1 , 0) onto ( T 5 3 , jf) and (0, 1) onto ( yf , ts).
Show that every line through the origin is rotated counterclockwise through the
angle 6 = arc cos T V This linear transformation is called a rotation. Find the
matrix representing this linear transformation with respect to the basis {(1,0),
(0,1)} , n
(/) A linear transformation maps (1 , 0) onto (f , f ) and (0, 1) onto (3 , 3). Show
that each point on the line 2x x + x 2 = 3c is mapped onto the single point (c, c).
The line x 1  x 2 = is left fixed. The only other line through the origin which
is mapped into itself is the line 2x l + x 2 = 0. This linear transformation is called
a projection onto the line x x  x 2 = parallel to the line 2x 1 + x 2 = 0. Find the
matrices representing this linear transformation with respect to the bases {(1 , 0),
(0, 1)} and {(1,1), (1, 2)}.
9. (Continuation) Describe the geometric effect of each of the linear transforma
tions of R 2 into itself represented by the matrices
(«)
"o r
(b)
"0 0"
(c)
"l r
1
1
"1 0"
~b 0"
r3 41
5 5
a 1
(e)
c
if)
4 3
5 5
id)
{Hint: In Exercise 7 we have shown that straight lines are mapped into straight
lines. We already know that linear transformations map the origin onto the origin.
Thus it is relatively easy to determine what happens to straight lines passing through
the origin. For example, to see what happens to the a^axis it is sufficient to see
what happens to the point (1,0). Among the transformations given appear a
rotation, a reflection, two projections, and one shear.)
10. (Continuation) For the linear transformations given in Exercise 9 find all
lines through the origin which are mapped onto or into themselves.
11. Let U = R 2 and V = R 3 and a be a linear transformation of U into V that
maps (1, 1) onto (0, 1, 2) and ( — 1, 1) onto (2, 1, 0). Determine the matrix that
represents a with respect to the bases A = {(1, 0), (0, 1)} in 8 = {(1, 0, 0), (0, 1,0),
(0,0, 1)} in R 3 . (Hint: (1> D ~ K"l. D = 0.0))
12. What is the effect of multiplying an n x n matrix A by an n x n diagonal
matrix Z>? What is the difference between AD and DAI
13. Let a and b be two numbers such that a ^ b. Find all 2 x 2 matrices A
such that
a
a
0"
b
A.
3  Nonsingular Matrices 45
14. Show that the matrix C = [a^bj] has rank one if not all a t and not all b t are
zero. {Hint: Use Theorem 1.12.)
15. Let a, b, c, and d be given numbers (real or complex) and consider the
function
ax + b
J ex + d
Let g be another function of the same form. Show that gf where gf{x) = g{f{x))
is a function that can also be written in the same form. Show that each of these
functions can be represented by a matrix in such a way that the matrix representing
gfis the product of the matrices representing £■ and/. Show that the inverse function
exists if and only if ad — be ^ 0. To what does the function reduce if ad — be =0?
16. Consider complex numbers of the form x + yi (where x and y are real
numbers and i 2 = — 1) and represent such a complex number by the duple {x, y)
in R 2 . Let a + bi be a fixed complex number. Consider the function / defined by
the rule
f{x + yi) = {a + bi){x + yi) = u + vi.
{a) Show that this function is a linear transformation of R 2 into itself mapping
{x, y) onto {u, v).
{b) Find the matrix representing this linear transformation with respect to the
basis {(1,0), (0,1)}.
(c) Find the matrix which represents the linear transformation obtained by using
c + di in place of a + bi. Compute the product of these two matrices. Do they
commute?
{d) Determine the complex number which can be used in place of a + bi to
obtain a transformation represented by this matrix product. How is this complex
number related to a + bi and c + dfl.
17. Show by example that it is possible for two matrices A and B to have the
same rank while A 2 and B 2 have different ranks.
3 I Nonsingular Matrices
Let us consider the case where U = V, that is, we are considering trans
formations of V into itself. Generally, a homomorphism of a set into itself
is called an endomorphism. We consider a fixed basis in V and represent
the linear transformation of V into itself with respect to that basis. In this
case the matrices are square or n x n matrices. Since the transformations
we are considering map Vinto itself any finite number of them can be iterated
in any order. The commutative law does not hold, however. The same
remarks hold for square matrices. They can be multiplied in any order but
46
Linear Transformations and Matrices I II
the commutative law does not hold. For example
"o r
"0 o~
"o r
_o o_
1_
o o_
"0 0"
"o r
"0 0"
_o L
_o o_
.o o_
The linear transformation that leaves every element of V fixed is the
identity transformation. We denote the identity transformation by 1,
the scalar identity. Clearly, the identity transformation is represented by
« the matrix / = [d u ] for any choice of the basis. Notice that I A = Al = A
for any n X n matrix A. I is called the identity matrix, or unit matrix, of
order n. If we wish to point out the dimension of the space we write /„ for
the identity matrix of order n. The scalar transformation a is represented
by the matrix al. Matrices of the form al are called scalar matrices.
Definition. A onetoone linear transformation a of a vector space onto
itself is called an automorphism. An automorphism is only a special kind of
isomorphism for which the domain and codomain are the same space. If
<x(a) = a, the mapping a 1 (a) = a is called the inverse transformation of a.
The rotations represented in Section 2 are examples of automorphisms.
* Theorem 3.1. The inverse a~ x of an automorphism a is an automorphism.
« Theorem 3.2 A linear transformation r of an ndimensional vector space
into itself is an automorphism if and only if it is of rank n ; that is, if and only if
it is an epimorphism.
» Theorem 3.3. A linear transformation a of an ndimensional vector space
into itself is an automorphism if and only if its nullity is 0, that is, if and only
if it is a monomorphism.
proof (of Theorems 3.1, 3.2, and 3.3). These properties have already
been established for isomorphisms. □
Since it is clear that transformations of rank less than n do not have
'inverses because they are not onto, we see that automorphisms are the
'only linear transformations which have inverses. A linear transformation
that has an inverse is said to be nonsingular or invertible; otherwise it is
said to be singular. Let A be the matrix representing the automorphism
a, and let A" 1 be the matrix representing the inverse transformation a~ x .
The matrix A~ X A represents the transformation a~ x a. Since cr x a is the
identity transformation, we must have A~ X A = I. But a is also the inverse
transformation of a~ x so that acr x = 1 and AA~ X = I. We shall refer to
A~ x as the inverse of A. A matrix that has an inverse is said to be non
singular or invertible. Only a square matrix can have an inverse.
3  Nonsingular Matrices
47
On the other hand suppose that for the matrix A there exists a matrix
B such that BA = I. Since / is of rank n, A must also be of rank n and, *
therefore, A represents an automorphism a. Furthermore, the linear
transformation which B represents is necessarily the inverse transformation
a" 1 since the product with a must yield the identity transformation. Thus
B = A* 1 . The same kind of argument shows that if C is a matrix such that
AC = I, then C = A~ x . Thus we have shown:
Theorem 3.4. If A and B are square matrices such that BA = /, then
AB = I. If A and B are square matrices such that AB = I, then BA = I.
In either case B is the unique inverse ofA.O
Theorem 3.5. If A and B are nonsingular, then (1) AB is nonsingular and
(AB) 1 = B^A' 1 , (2) A' 1 is nonsingular and (A' 1 ) 1 = A, (3) for a ^ 0,
aA is nonsingular and (aA) 1 = a~ 1 A~ 1 .
proof. In view of the remarks preceding Theorem 3.4 it is sufficient in
each case to produce a matrix which will act as a left inverse.
(1) (B^A^iAB) = B 1 (A~ 1 A)B = B~ X IB = B~ X B = I.
(2) AA' 1 = I.
(3) (a^A^iaA) = (a^aXA^A) = /. □
Theorem 3.6. If A is nonsingular, we can solve uniquely the equations
XA = B and AY = Bfor any matrix B of the proper size, but the two solutions
need not be equal.
proof. Solutions exist since {BA~ X )A = B{A~ X A) = B and A(A~ X B) =
(AA _1 )B = B. The solutions are unique since for any C having the property
that CA — B we have C = CAA~ X = BA~ X , and similarly with any solution
of ,4 7= B. a
As an example illustrating the last statement of the theorem, let
A =
Then
X = BA 1 =
1
2"
1_
»
"1
2
2
3
A~ x =
and
B =
Y = A~ X B =
I 0"
I 1_
3
2
2
1
We add the remark that for nonsingular A, the solution of XA = B
exists and is unique if B has n columns, and the solution of AY = B exists
and is unique if B has n rows. The proof given for Theorem 3.6 applies
without change.
48
Linear Transformations and Matrices I II
Theorem 3.7. The rank of a (not necessarily square) matrix is not changed
by multiplication by a nonsingular matrix.
proof. Let A be nonsingular and let B be of rank p. Then by Theorem
2.1 AB is of rank r < p, and A^iAB) = B is of rank p < r. Thus r = p.
The proof that BA is of rank p is similar. □
Theorem 1.14 states the corresponding property for linear transformations.
The existence or nonexistence of the inverse of a square matrix depends
on the matrix itself and not on whether it represents a linear transformation
of a vector space into itself or a linear transformation of one vector space
into another. Thus it is convenient and consistent to extend our usage of the
term "nonsingular" to include isomorphisms. Accordingly any square
matrix with an inverse is nonsingular.
Let U and V be vector spaces of dimension n over the field F. Let A =
{aj, . . . , a n } be a basis of U and 8 = {(S x , . . . , B n } be a basis of V. If
£ = XLi^t i s an y vector in U we can define <r(£) to be ^Li^f ^ * s
easily seen that <r is an isomorphism and that £ and cr(£) are both repre
sented by (#!, . . . , x n ) e F n . Thus any two vector spaces of the same
dimension over F are isomorphic. As far as their internal structure is con
cerned they are indistinguishable. Whatever properties may serve to dis
tinguish them are, by definition, not vector space properties.
EXERCISES
1 . Show that the inverse of
"1 2 3"
A = 2 3 4
3 4 6
2. Find the square of the matrix
is
A' 1 =
2
1
3
2
1"
2
1
2
2
1
2"
1
2
What is the inverse of Al (Geometrically, this matrix represents a 180° rotation
about the line containing the vector (2, 1, 1). The inverse obtained is therefore
not surprising.)
3. Compute the image of the vector (1, —2, 1) under the linear transformation
represented by the matrix
"1 2 3"
A =
Show that A cannot have an inverse.
2 3 4
1 2
3  Nonsingular Matrices
4. Since
T ll X 12
49
we can find the inverse of
3 1
5 2
" 3 1
_5 2
ix 1:l 5cc 12
11 ^12 ~"~ 11 ' ^^12
JX 21 J^22 "^21 "r" •^ a '22
by solving the equations
= 1
=
•'• C 21 5#22 =
X %\ "■ ■^" c 22 = 1
Solve these equations and check your answer by showing that this gives the inverse
matrix.
We have not as yet developed convenient and effective methods for obtaining
the inverse of a given matrix. Such methods are developed later in this chapter
and in the following chapter. If we know the geometric meaning of the matrix,
however, it is often possible to obtain the inverse with very little work.
5. The matrix
represents a rotation about the origin through the angl&
6 = arc cos f. What rotation would be the inverse of this rotation ? What matrix
would represent this inverse rotation? Show that this matrix is the inverse of the
given matrix.
• r ° _i i
6. The matrix represents a reflection about the line x, + x 2 = 0.
1 OJ
What operation is the inverse of this reflection? What matrix represents the
inverse operation? Show that this matrix is the inverse of the given matrix.
n n
7. The matrix represents a shear. The inverse transformation is also a
shear. Which one? What matrix represents the inverse shear? Show that this
matrix is the inverse of the given matrix.
8. Show that the transformation that maps (x x , x 2 , x 3 ) onto (x 3 , — x x , x 2 ) is an
automorphism of F 3 . Find the matrix representing this automorphism and its
inverse with respect to the basis {(1,0, 0), (0, 1, 0), (0, 0, 1)}.
9. Show that an automorphism of a vector space maps every subspace onto a
subspace of the same dimension.
10. Find an example to show that there exist nonsquare matrices A and B
such that AB — I. Specifically, show that there is an m x n matrix A and an
n x m matrix B such that AB is the m x m identity. Show that BA is not the
n x n identity. Prove in general that if m ^ n, then AB and BA cannot both be
identity matrices.
50 Linear Transformations and Matrices  II
4 I Change of Basis
We have represented vectors and linear transformations as /ztuples and
matrices with respect to arbitrary but fixed bases. A very natural question
arises: What changes occur in these representations if other choices for
bases are made? The vectors and linear transformations have meaning
independent of any particular choice of bases, independent of any coordinate
systems, but their representations are entirely dependent on the bases chosen.
Definition. Let A = {a l9 . . . , a„} and A' = {x' v . . . , ol'J be bases of the
vector space U. In a typical "change of basis" situation the representations
of various vectors and linear transformations are known in terms of the
basis A, and we wish to determine their representations in terms of the
basis A'. In this connection, we refer to A as the "old" basis and to A' as
the "new" basis. Each ex.'. is expressible as a linear combination of the
elements of A; that is,
*; = i>««i. P (4.i)
The associated matrix P = [p i0 \ is called the matrix of transition from the
basis A to the basis A\
The columns of P are the ^tuples representing the new basis vectors in
terms of the old basis. This simple observation is worth remembering as
it is usually the key to determining P when a change of basis is made. Since
the columns of P are the representations of the basis A' they are linearly
independent and P has rank n. Thus P is nonsingular.
Now let £ = 2"=i x iV*i be an arbitrary vector of U and let £ = 2> =1 x\ix i ~ s
be the representation of £ in terms of the basis A'. Then <>
n n / n \
= 2 llPiri W ( 4  2 >
i=i y=i /
Since the representation of £ with respect to the basis A is unique we see
that x t = 2j l =i/ > tf a: i Notice that the rows of P are used to express the
old coordinates of I in terms of the new coordinates. For emphasis and
contradistinction, we repeat that the columns of P are used to express the
new basis vectors in terms of the old basis vectors.
Let X = (x x , . . . , x n ) and X' = (x[, . . . , x^) be n x 1 matrices representing
the vector I with respect to the bases A and A'. Then the set of relations
{ x i = ^j=\Pio x 'j) can ^ e written as the single matric equation
X = PX'. (4.3)
4  Change of Basis 51
Now suppose that we have a linear transformation a of U into V and
that A = [a i} ] is the matrix representing a with respect to the bases A in
U and B = {ft, . . . , ftj in V. We shall now determine the representation
of a with respect to the bases A' and B.
n < ' f n I m v
k=l k=l \i=l /
to J n \
"\" h ^c *, ^/,.,U * N ■■■■■' ■
«i x ( . •
= 1<A V^o f\h ■ (4.4)
1=1
Since 6 is a basis, a' i:j = 2jL x a uPki and tne matrix ^f' = [a' i3 ] representing
a with respect to the bases A' and 6 is related to A by the matric equation
A' = AP. (4.5)
This relation can also be demonstrated in a slightly different way. For
an arbitrary f = 2» =1 a^a, e U let <r(£) = £™ r */,&• Then we have
Y.= AX = ,4 (/>*') = (AP)A". (4.6)
Thus v4P is a matrix representing c with respect to the bases A' and B. Since
the matrix representing a is uniquely determined by the choice of bases we
have A' = AP.
Now consider the effect of a change of basis in the image space V. Thus
let B be replaced by the basis B' = {ft, . . . , ftj. Let = fo w ] be the
matrix of transition from B to B', that is, ft = 2™ ^ft. Then if >4" =
[a".] represents a with respect to the bases A and B' we have
TO TO / TO \
<r(a,) = 2X;& = 2 a'ki ( 2 0*A)
TO / TO v TO
= 2(I«tt*wU = 2««A. (4.7)
1=1 \fc=l / i=l
Since the representation of cr(a,) in terms of the basis B is unique we see
that A = QA", or
4" = g 1 ^. (4.8)
Combining these results, we see that, if both changes of bases are made at
once, the new matrix representing a is Q~ X AP.
As in the proof of Theorem 1.6 we can choose a new basis A' = {04, . . . , <x.' n }
of U such that the last v = n — p basis elements form a basis of K{a). Since
M04), . . . , <r(<Xp)} is a basis of a(U) and is linearly independent in V, it can
52
Linear Transformations and Matrices I II
be extended to a basis B' of V. With respect to the bases A' and 8' we have
or(a') = ft for/" < p while cr(a^) = for/ > p. Thus the new matrix Q~ X AP
representing a is of the form
p columns
10
1
p rows
m — p rows
v columns
Thus we have
Theorem 4.1. If A is any m X n matrix of rank p, there exist a non
singular n x n matrix P and a nonsingular m x m matrix Q such that
A' = Q~ X AP has the first p elements of the main diagonal equal to 1, and
all other elements equal to zero. □
When A and B are unrestricted we can always obtain this relatively simple
representation of a linear transformation by a proper choice of bases.
More interesting situations occur when A and B are restricted. Suppose,
for example, that we take U = V and A = B. In this case there is but one
basis to change and but one matrix of transition, that is, P = Q. In this
case it is not possible to obtain a form of the matrix representing a as simple
as that obtained in Theorem 4.1. We say that any two matrices representing
the same linear transformation or of a vector space V into itself are similar.
This is equivalent to saying that two matrices A and A' are similar if and
only if there exists a nonsingular matrix of transition P such that A' =
P X AP. This case occupies much of our attention in Chapters III and V.
EXERCISES
1. In P 3 , the space of polynomials of degree 2 or smaller with coefficients in
F, let A ={\,x,x 2 }.
A' = { Pl {x) = x 2 + x + 1 , p 2 (x) =x 2 x 2, p 3 (x) = x 2 + x  1}
is also a basis. Find the matrix of transition from A to A'.
5 [ Hermite Normal Form 53
2. In many of the uses of the concepts of this section it is customary to take
A = {a. i \x i = (d n , 6 i2 , ... , d in )} as the old basis in R n . Thus, in R 2 let A =
{(1, 0), (0, 1)} and A' = {(£, V3/2), ( V3/2, £)}. Show that
P =
\ V3/2
V3/2 i
is the matrix of transition from A to A'.
3. (Continuation) With A' and A as in Exercise 2, find the matrix of transition R
from A' to A. (Notice, in particular, that in Exercise 2 the columns of P are the
components of the vectors in A' expressed in terms of basis A, whereas in this exercise
the columns of R are the components of the vectors in A expressed in terms of the
basis A'. Thus these two matrices of transition are determined relative to different
bases.) Show that RP = I.
4. (Continuation) Consider the linear transformation of o of R 2 into itself which
maps
(1,0) onto (£, V3/2)
(0,1) onto (V3/2,).
Find the matrix A that represents a with respect to the basis A.
You should obtain A = P. However, A and P do not represent the same thing.
To see this, let £ = (x lt x 2 ) be an arbitrary vector in R 2 and compute a(£) by means
of formula (2.9) and the new coordinates of £ by means of formula (4.3).
A little reflection will show that the results obtained are entirely reasonable.
The matrix A represents a rotation of the real plane counterclockwise through an
angle of w/3. The matrix P represents a rotation of the coordinate axes counter
clockwise through an angle of w/3. In the latter case the motion of the plane
relative to the coordinate axes is clockwise through an angle of 77/3.
5. In R 3 let A = {(1,0, 0), (0, 1,0), (0, 0, 1)} and let A' = {(0, 1, 1), (1,0, 1),
(1,1, 0)}. Find the matrix of transition P from A to A' and the matrix of transition
P 1 from A' to A.
6. Let A, 8, and C be three bases of V. Let P be the matrix of transition from A
to 6 and let Q be the matrix of transition from 8 to C. Is PQ or QP the matrix of
transition from A to C? Compare the order of multiplication of matrices of transi
tion and matrices representing linear transformation.
7. Use the results of Exercise 6 to resolve the question raised in the parenthetical
remark of Exercise 3, and implicitly assumed in Exercise 5. If P is the matrix of
transition from A to A' and Q is the matrix of transition from A' to A, show that
PQ =/.
5 I Hermite Normal Form
We may also ask how much simplification of the matrix representing a
linear transformation a of U into V can be effected by a change of basis in
54
Linear Transformations and Matrices I II
V alone. Let A = {a l5 . . . , a n } be the given basis in U and let U k = (a l5 . . . ,
a fc ). The subspaces 0(U k ) of V form a nondecreasing chain of subspaces with
«(U k i) <= <y(U k ) and o"(U n ) = cr(U). Since a(U k ) = a^) + (a(oi k )) we see
from Theorem 4.8 of Chapter I that dim a(U k ) < dim o^L/^) + 1 ; that is,
the dimensions of the o"(U fc ) do not increase by more than 1 at a time as k
increases. Since dim (r(U n ) = p, the rank of a, an increase of exactly 1
must occur p times. For the other times, if any, we must have dim c(U k ) =
dim ^(L/fc.i) and hence cr(U k ) = cf(U k _^). We have an increase by 1 when
cf(ce. k ) <£ ^(C^i) and no increase when a{a. k ) e oiU^).
Let k lf k 2 , . . . , k p be those indices for which o(a fc .) <£ a(U k 1 ). Let
ft = ^K) Since ft £ er^) = (ft, . . . , ft_ x >, the set {ft, . .'. , ft} is
linearly independent (see Theorm 2.3, Chapter 12). Since {ft, . . . , ft} <=
<r(t/) and cr(l/) is of dimension p, {ft, . . . , ft} is a basis of a(U). This set
can be extended to a basis B' of V. Let us now determine the form of the
matrix A' representing a with respect to the bases A and 8'.
Since o(ct. k ^ = ft, column k t has a 1 in row i and all other elements of
this column are O's. For k t <j < k i+1 , <r(oc 3 ) e <r(C fc .) so that column j
has O's below row /. In general, there is no restriction on the elements of
column j in the first i rows. A' thus has the form
column
column
*1
/c 2
•
•
1
"l,fcl+l •
..
r
"l,fc 2 +l
•
•
• • 1
a 2,fc 2 +l
•
•
•
(5.1)
Once A and a are given, the k i and the set {ft, . . . , ft} are uniquely
determined. There may be many ways to extend this set to the basis 6',
but the additional basis vectors do not affect the determination of A' since
every element of a{U) can be expressed in terms of {ft, ... , ft} alone. Thus
A' is uniquely determined by A and a.
Theorem 5.1. Given any m x n matrix A of rank p, there exists a non
singular m x m matrix Q such that A' = Q~ X A has the following form:
(1) There is at least one nonzero element in each of the first p rows of A' ,
and the elements in all remaining rows are zero.
5  Hermite Normal Form 55
(2) The first nonzero element appearing in row i (i < p) is a 1 appearing
in column k t , where k x < k 2 < • • • < k p .
(3) In column k t the only nonzero element is the 1 in row i.
The form A' is uniquely determined by A.
proof. In the applications of this theorem that we wish to make A is
usually given alone without reference to any bases A and 6, and often without
reference to any linear transformation a. We can, however, introduce any
two vector spaces U and V of dimensions n and m over F and let A be any
basis of U and 8 be any basis of V. We can consider A as defining a linear
transformation a of U into V with respect to the bases A and 8. The discussion
preceding Theorem 5.1 shows that there is at least one nonsingular matrix
Q such that Q _1 A satisfies conditions (1), (2), and (3).
Now suppose there are two nonsingular matrices Q x and Q 2 such that
Ql x A = A' x and Q 2 X A = A 2 both satisfy the conditions of the theorem.
We wish to conclude that A[ = A' 2 . No matter how the vector spaces U
and V are introduced and how the bases A and 8 are chosen we can regard
Q x and Q 2 as matrices of transition in V. Thus A[ represents a with respect
to bases A and 8^ and A' 2 represents a with respect to bases A and & 2 . But
condition (3) says that for i < p the rth basis element in both 8^ and B' 2 is
<7(<x fc .). Thus the first p elements of B[ and B' 2 are identical. Condition (1) says
that the remaining basis elements have nothing to do with determining the
coefficients in A[ and A' 2 . Thus A' x = A 2 . □
We say that a matrix satisfying the conditions of Theorem 5.1 is in
Hermite normal form. Often this form is called a rowechelon form. And
sometimes the term, Hermite normal form, is reserved for a square matrix
containing exactly the numbers that appear in the form we obtained in
Theorem 5.1 with the change that row /' beginning with a 1 in column k t
is moved down to row k t . Thus each nonzero row begins on the main
diagonal and each column with a 1 on the main diagonal is otherwise zero.
In this text we have no particular need for this special form while the form
described in Theorem 5.1 is one of the most useful tools at our disposal.
The usefulness of the Hermite normal form depends on its form, and
the uniqueness of that form will enable us to develop effective and con
venient short cuts for determining that form.
Definition. Given the matrix A, the matrix A T obtained from A by inter
changing rows and columns in A is called the transpose of A. If A T = [a'^],
the element a' tj appearing in row i column j of A T is the element a H appear
ing in row j column / of A. It is easy to show that (AB) T = B T A T . (See
Exercise 4.)
Proposition 5.2. The number of linearly independent rows in a matrix is
equal to the number of linearly independent columns.
56
Linear Transformations and Matrices I II
proof. The number of linearly independent columns in a matrix A is its
rank p. The Hermite normal form A' = Q~ X A corresponding to A is also
of rank p. For A' it is obvious that the number of linearly independent rows
in A' is also equal to p, that is, the rank of (A') T is p. Since Q T is non
singular, the rank of A T = (QA') T = (A') T Q T is also p. Thus the number
of linearly independent rows in A is p. □
(a)
(b)
EXERCISES
1. Which of the following matrices are in Hermite normal form?
"01001"
10 1
10
0_
"00204"
110 3
12
0_
"i o o o r
10 1
11
0_
"0101001"
10 10
10
0_
"10 10 1"
110
oooio
lj
2. Determine the rank of each of the matrices given in Exercise 1.
3. Let a and r be linear transformations mapping R 3 into R 2 . Suppose that for
a given pair of bases A for R 3 and B for R 2 , a and t are represented by
(c)
(d)
(e)
A =
T 1 0'
1
and
B =
1 1"
1
6  Elementary Operations and Elementary Matrices
57
respectively. Show that there is no basis 8' of R 2 such that B is the matrix represent
ing a with respect to A and 8'.
4. Show that
(a) (A + B) T = A T + B T ,
(b) (AB) T = B T A T ,
(c) {A^)T = (AT)\
6 I Elementary Operations and Elementary Matrices
Our purpose in this section is to develop convenient computational
methods. We have been concerned with the representations of linear
transformations by matrices and the changes these matrices undergo when
a basis is changed. We now show that these changes can be effected by
elementary operations on the rows and columns of the matrices.
We define three types of elementary operations on the rows of a matrix A.
Type I : Multiply a row of A by a nonzero scalar.
Type II : Add a multiple of one row to another row.
Type III : Interchange two rows.
Elementary column operations are defined in an analogous way.
From a logical point of view these operations are redundant. An opera
tion of type III can be accomplished by a combination of operations of
types I and II. It would, however, require four such operations to take the
place of one operation of type III. Since we wish to develop convenient
computational methods, it would not suit our purpose to reduce the number
of operations at our disposal. On the other hand, it would not be of much
help to extend the list of operations at this point. The student will find that,
with practice, he can combine several elementary operations into one step.
For example, such a combined operation would be the replacing of a row
by a linear combination of rows, provided that the row replaced appeared
in the linear combination with a nonzero coefficient. We leave such short
cuts to the student.
An elementary operation can also be accomplished by multiplying A on
the left by a matrix. Thus, for example, multiplying the second row by the
scalar c can be effected by the matrix
1 ••• (T
E 2 (c) =
c •••
1 •••
(6.1)
58
Linear Transformations and Matrices I II
The addition of k times the third row to the first row can be effected by the
matrix
E sl (k) =
"1 k •
• 0"
10
•
1
•
(6.2)
_0 • • • 1
The interchange of the first and second rows can be effected by the matrix
~0 1 • • • 0"
1 •••
1 •••
•£l9
(6.3)
_o o o ••• :
These matrices corresponding to the elementary operations are called
elementary matrices. These matrices are all nonsingular and their inverses
are also elementary matrices. For example, the inverses of E 2 (c), E 31 (k), and
E 12 are respectively E^c' 1 ), E 31 (—k), and E 12 
Notice that the elementary matrix representing an elementary operation is
the matrix obtained by applying the elementary operation to the unit matrix.
Theorem 6.1. Any nonsingular matrix A can be written as a product of
elementary matrices.
proof. At least one element in the first column is nonzero or else A
would be singular. Our first goal is to apply elementary operations, if
necessary, to obtain a 1 in the upper lefthand corner. If a n = 0, we can
interchange rows to bring a nonzero element into that position. Thus we
may as well suppose that a lx ^ 0. We can then multiply the first row by
an" 1 . Thus, to simplify notation, we may as well assume that a lx = 1.
We now add —a a times the first row to the rth row to make every other
element in the first column equal to zero.
The resulting matrix is still nonsingular since the elementary operations
applied were nonsingular. We now wish to obtain a 1 in the position of
element a 22 . At least one element in the second column other than a 12
6  Elementary Operations and Elementary Matrices 59
is nonzero for otherwise the first two columns would be dependent. Thus
by a possible interchange of rows, not including row 1, and multiplying the
second row by a nonzero scalar we can obtain a 22 = 1 We now add — a iZ
times the second row to the /th row to make every other element in the second
column equal to zero. Notice that we also obtain a in the position of a 12
without affecting the 1 in the upper lefthand corner.
We continue in this way until we obtain the identity matrix. Thus if
E lt E 2 , . . . ,E r are elementary matrices representing the successive elementary
operations, we have
I = E r E 2 E l A, ?'■ M V ' : 
or (6.4)
A = E?E^E~\n T; ^ p  r .
In Theorem 5.1 we obtained the Hermite normal form A' from the matrix
A by multiplying on the left by the nonsingular matrix Q _1 . We see now
that Q~ x is a product of elementary matrices, and therefore that A can be
transformed into Hermite normal form by a succession of elementary row
operations. It is most efficient to use the elementary row operations directly
without obtaining the matrix Q~ x .
We could have shown directly that a matrix could be transformed into
Hermite normal form by means of elementary row operations. We would
then be faced with the necessity of showing that the Hermite normal form
obtained is unique and not dependent on the particular sequence of oper
ations used. While this is not particularly difficult, the demonstration is
uninteresting and unilluminating and so tedious that it is usually left as an
"exercise for the reader." Uniqueness, however, is a part of Theorem 5.1,
and we are assured that the Hermite normal form will be independent of
the particular sequence of operations chosen. This is important as many
possible operations are available at each step of the work, and we are free
to choose those that are most convenient.
Basically, the instructions for reducing a matrix to Hermite normal form
are contained in the proof of Theorem 6.1. In that theorem, however, we
were dealing with a nonsingular matrix and thus assured that we could
at certain steps obtain a nonzero element on the main diagonal. For a
singular matrix, this is not the case. When a nonzero element cannot be
obtained with the instructions given we must move our consideration to the
next column.
In the following example we perform several operations at each step to
conserve space. When several operations are performed at once, some
care must be exercised to avoid reducing the rank. This may occur, for
example, if we subtract a row from itself in some hidden fashion. In this
example we avoid this pitfall, which can occur when several operations of
f\" fi
60
Linear Transformations and Matrices I II
type III are combined, by considering one row as an operator row and adding
multiples of it to several others.
Consider the matrix
4
3
2
1
4
• 5
4
3
1
4
2
2
1
2
3
11
6
4
1
11
as an example.
According to the instructions for performing the elementary row oper
ations we should multiply the first row by J. To illustrate another possible
way to obtain the "1" in the upper left corner, multiply row 1 by —1 and
add row 2 to row 1 . Multiples of row 1 can now be added to the other rows
to obtain
1
1
1
1
2
1
4
1
2
3
5
7
1
11
Now, multiply row 2 by —1 and add appropriate multiples to the other
rows to obtain
Finally, we obtain
1
1
1
4
1
2
1
4
1
2
3
3
6
9
"l
1
1
1
3
2
1
2
3
which is the Hermite normal form described in Theorem 5.1. If desired,
Q~ x can be obtained by applying the same sequence of elementary row
operations to the unit matrix. However, while the Hermite normal form
is necessarily unique, the matrix Q~ x need not be unique, as the proof of
Theorem 5.1 should show.
6  Elementary Operations and Elementary Matrices
61
Rather than trying to remember the sequence of elementary operations
used to reduce A to Hermite normal form, it is more efficient to perform
these operations on the unit matrix at the same time we are operating on
A. It is suggested that we arrange the work in the following way:
3 2
4 3
2 1
6 4
1 1
1 2 1
1 2
5 7
1
1 2
1
1
2
1
1
1
1
4
4
3
11
1
4 5
3 2
1
4
2
2
6
1
3
2
11
4
4
3
11 11
4
5
2
9 14
1 2
2 1
3 2
8
In the end we obtain
0i =
2
1
2
1
2
3
3
4
2
9
1
2
3
1
2
1
3
1
1
1
o"
2
1
3
1
= [A, I]
Verify directly that Q~ X A is in Hermite normal form.
If A were nonsingular, the Hermite normal form obtained would be the
identity matrix. In this case Q~ x would be the inverse of A. This method
of finding the inverse of a matrix is one of the easiest available for hand
computation. It is the recommended technique.
62
Linear Transformations and Matrices I II
EXERCISES
1. Elementary operations provide the easiest methods for determining the rank
of a matrix. Proceed as if reducing to Hermite normal form. Actually, it is not
necessary to carry out all the steps as the rank is usually evident long before the
Hermite normal form is obtained. Find the ranks of the following matrices :
(a)
(b)
(c)
'1 2 3
4 5 6
7 8 9
' 1
1
2 3
"0
1
2"
1
3
2
3
2. Identify the elementary operations represented by the following elementary
matrices :
(a)
(b)
(c)
1
2
"0
r
1
1
"i
0"
2
1
3. Show that the product
"1 0"
_ 1 0"
"1
r
"1 0"
1
1 1
i
i i
is an elementary matrix. Identify the elementary operations represented by each
matrix in the product.
4. Show by an example that the product of elementary matrices is not necessarily
an elementary matrix.
7  Linear Problems and Linear Equations
63
5. Reduce each of the following matrices to Hermite normal form.
(a)
(b)
"2 1
2 1
3 2'
5 2
1 1 1
12 3 3
2 10
2 2 2 1
1
10 6"
2 3
5 5
113 2 5 2_
6. Use elementary row operations to obtain the inverses of
(a) I" 3 11
—5 2 , and
(b) Tl 2 3
2 3 4
3 4 6.
7. (a) Show that, by using a sequence of elementary operations of type II only,
any two rows of a matrix can be interchanged with one of the two rows multiplied
by —1. (In fact, the type II operations involve no scalars other than ±1.)
(b) Using the results of part (a), show that a type III operation can be obtained
by a sequence of type II operations and a single type I operation.
(c) Show that the sign of any row can be changed by a sequence of type II
operations and a single type III operation.
8. Show that any matrix A can be reduced to the form described in Theorem 4.1
by a sequence of elementary row operations and a sequence of elementary column
operations.
7 I Linear Problems and Linear Equations
For a given linear transformation a of U into V and a given p e V the
problem of finding any or all gU for which o(£) = /S is called a linear
problem. Before providing any specific methods for solving such problems,
let us see what the set of solutions should look like.
If jS <£ cf(U), then the problem has no solution.
If /? g a(U), the problem has at least one solution. Let £ °e one such
solution. We call any such  a. particular solution. If I is any other solution,
then <r(£   ) = <*(£ )  <?(£o) = £  P = so that f  f is in the kernel
of a. Conversely, if £ — £„ is in the kernel of c then a (!) = <r( + £ — £ ) =
or(£ o ) + o( _ £ ) — £ + o = /S so that f is a solution. Thus the set of all
solutions of (r(!) = /3 is of the form
{f } + K{a). (7.1)
64 Linear Transformations and Matrices  II
Since {£ } contains just one element, there is a onetoone correspondence
between the elements of K{a) and the elements of {£ } + K{a). Thus the
size of the set of solutions can be described by giving the dimension of K(a).
The set of all solutions of the problem <r(£) = is not a subspace of U unless
/8 = 0. Nevertheless, it is convenient to say that the set is of dimension v,
the nullity of a.
Given the linear problem <r() = 0, the problem <r(f) = is called the
associated homogeneous problem. The general solution is then any particular
solution plus the solution of the associated homogeneous problem. The
solution of the associated homogeneous problem is the kernel of a.
Now let o be represented by the m x n matrix A = [a tj ], be represented
by B = (b u ... , bj, and £ by X = (x u . . . , x n ). Then the linear problem
o(g) = f} becomes
AX = B (7.2)
in matrix form, or
!>„*, = &„ (i=l,...,m) (7.3)
in the form of a system of linear equations.
Given A and 5, the augmented matrix [A, B] of the system of linear
equations is defined to be
[A, B] =
«n   " a ln b x
(7.4)
Theorem 7.1. The system of simultaneous linear equations AX = B has a
solution if and only if the rank of A is equal to the rank of the augmented
matrix [A, B]. Whenever a solution exists, all solutions can be expressed in
terms of v = n — p independent parameters , where p is the rank of A.
proof. We have already seen that the linear problem <r() = 8 has a
solution if and only if e o(U). This is the case if and only if is linearly
dependent on MoO, . . . , cr(a n )}. But this is equivalent to the condition
that B be linearly dependent on the columns of A. Thus adjoining the
column of b/s to form the augmented matrix must not increase the rank.
Since the rank of the augmented matrix cannot be less than the rank of A
we see that the system has a solution if and only if these two ranks are equal.
Now let Q be a nonsingular matrix such that Q^A = A' is in Hermite
normal form. Any solution of AX = B is also a solution of A'X = Q~ X AX =
Q~ X B = B'. Conversely, any solution of A'X = B' is also a solution of
AX = QA'X = QB' = B. Thus the two systems of equations are equivalent.
7  Linear Problems and Linear Equations
65
Now the system A'X = B' is particularly easy to solve since the variable
x k . appears only in the ith equation. Furthermore, nonzero coefficients
appear only in the first p equations. The condition that /? e a(U) also
takes on a form that is easily recognizable. The condition that B' be ex
pressible as a linear combination of the columns of A' is simply that the
elements of B' below row p be zero. The system A'X = B' has the form
+ ^l.fcj+i^fci+i +" • • + ai,fc 2+ i#fc 2+ i +
+ a'
2,fc 2 +l^fc 2 +l
vCt.
(7.5)
Since each x k . appears in but one equation with unit coefficient, the remaining
n — p unknowns can be given values arbitrarily and the corresponding
values of the x k . computed. The n — p unknowns with indices not the k t
are the n — p parameters mentioned in the theorem. □
As an example, consider the system of equations:
5x x + 4x 2 + 3# 3 — x t = 4
\\x x + 6x 2 + 4x 3 + # 4 = 11.
The augmented matrix is
' A 3 21 4
5 4 314
2 2 1 2 3
11 6 4 1 11
This is the matrix we chose for an example in the previous section. There
we obtained the Hermite normal form
1
1
1
1
3
2
1
2
3
Thus the system of equations A'X = B' corresponding to this augmented
matrix is
X \ + #4=1
x % + 2z 4 = —3.
66 Linear Transformations and Matrices  II
It is clear that this system is very easy to solve. We can take any value
whatever for x 4 and compute the corresponding values for x lt x 2 , and x z .
A particular solution, obtained by taking x 4 = 0, is X = (1, 2, 3, 0).
It is more instructive to write the new system of equations in the form
x± = l — x 4
%2 == *• i *%&
X z i= — J — Zx 4
%4 = X^
In vector form this becomes
(a?!, x 2 , x z , Xi ) = (1,2, 3, 0) + * 4 (l, 3, 2, 1).
We can easily verify that (1,3, 2, 1) is a solution of the associated
homogeneous problem. In fact, {(1, 3, 2, 1)} is a basis for the kernel,
and * 4 (— 1, 3, —2, 1), for an arbitrary z 4 , is a general element of the kernel.
We have, therefore, expressed the general solution as a particular solution
plus the kernel.
The elementary row operations provide us with the recommended technique
for solving simultaneous linear equations by hand. This application is the
principal reason for introducing elementary row operations rather than
column operations.
Theorem 7.2. The equation AX = B fails to have a solution if and only if
there exists a onerow matrix C such that CA = and CB = 1.
proof. Suppose the equation AX = B has a solution and a C exists
such that CA = and CB = 1 . Then we would have = (CA)X = C(AX) =
CB = 1 , which is a contradiction.
On the other hand, suppose the equation AX = B has no solution. By
Theorem 7.1 this implies that the rank of the augmented matrix [A, B] is
greater than the rank of A. Let Q be a nonsingular matrix such that
<2 1 l>4>#] is in Hermite normal form. Thenif pis therankof,4, the (p + l)st
row of Q~ X [A, B] must be all zeros except for a 1 in the last column. If C
is the (p + l)st row of Q~ x this means that
C[A,B] = [0 ••• 1],
or
CA = and CB = 1. □
This theorem is important because it provides a positive condition for a
negative conclusion. Theorem 7.1 also provides such a positive condition
and it is to be preferred when dealing with a particular system of equations.
But Theorem 7.2 provides a more convenient condition when dealing with
systems of equations in general.
Although the sytems of linear equations in the exercises that follow are
written in expanded form, they are equivalent in form to the matric equation
7  Linear Problems and Linear Equations 67
AX = B. From any linear problem in this set, or those that will occur later,
it is possible to obtain an extensive list of closely related linear problems
that appear to be different. For example, if AX = B is the given linear
problem with A an m x n matrix and Q is any nonsingular m x m matrix,
then A'X = B' with A' = QA and B' = QB is a problem with the same set of
solutions. If Pis a nonsingular n x n matrix, then A"X" = B where A" = AP
is a problem whose solution X" is related to the solution X of the original
problem by the condition X" = P X X.
For the purpose of constructing related exercises of the type mentioned,
it is desirable to use matrices P and Q that do not introduce tedious numerical
calculations. It is very easy to obtain a nonsingular matrix P that has only
integral elements and such that its inverse also has only integral elements.
Start with an identity matrix of the desired order and perform a sequence of
elementary operations of types II and III. As long as an operation of type I is
avoided, no fractions will be introduced. Furthermore, the inverse opera
tions will be of types II and III so the inverse matrix will also have only
integral elements.
For convenience, some matrices with integral elements and inverses with
integral elements are listed in an appendix. For some of the exercises that
are given later in this book, matrices of transition that satisfy special con
ditions are also needed. These matrices, known as orthogonal and unitary
matrices, usually do not have integral elements. Simple matrices of these
types are somewhat harder to obtain. Some matrices of these types are also
listed in the appendix.
EXERCISES
1. Show that {(1,1,1, 0), (2, 1, 0, 1)} spans the subspace of all solutions of the
system of linear equations
3x x — 2x 2 — x z — 4# 4 =
x i + x 2 — 2x 3 — 3x 4 = 0.
2. Find the subspace of all solutions of the system of linear equations
x x + 2x 2 — 3*3 +x i =
3x x — x 2 + 5x 3 — x i =
2x x + x 2 a; 4 = 0.
3. Find all solutions of the following two systems of nonhomogeneous linear
equations.
(a) x x + 3x 2 + 5x 3 — 2x 4 = 11
3*! — 2x 2 — 7x 3 + 5* 4 =
2x x + x 2 + x 4 = 7,
(b) x x + 3x 2 + 2x 3 + 5x 4 = 10
3x x — 2x 2 — 5x 3 + 4x i = —5
2x x + x 2  x 3 + 5x 4 = 5.
68 Linear Transformations and Matrices  II
4. Find all solutions of the following system of nonhomogeneous linear equations
ZtXt *^2 "^3 — ■*■
Ju i ~ ~ JC o ~T~ £•& o — — "" ~~ ^
^tJU% J&q i~ *& q — — *" "" J
5. Find all solutions of the system of equations,
7x 1 + 3x 2 + 21^3 — \3x i + x 5 = —14
lOo^ + 3x 2 + 3Cte 3 — 16x 4 + £ 5 = —23
7x 1 + 2« 2 + 21x 3 — llx 4 + x 5 = —16
9^! + 3z 2 + 27ic 3  15a; 4 + x 5 = 20.
6. Theorem 7.1 states that a necessary and sufficient condition for the existence
of a solution of a system of simultaneous linear equations is that the rank of the
augmented matrix be equal to the rank of the coefficient matrix. The most efficient
way to determine the rank of each of these matrices is to reduce each to Hermite
normal form. The reduction of the augmented matrix to normal form, however,
automatically produces the reduced form of the coefficient matrix. How, and
where? How is the comparison of the ranks of the coefficient matrix and the
augmented matrix evident from the appearance of the reduced form of the aug
mented matrix ?
7. The differential equation d 2 y/dx 2 + Ay = sin x has the general solution
y = C x sin 2x + C 2 cos 2x +  sin x. Identify the associated homogeneous prob
lem, the solution of the associated homogeneous problem, and the particular solu
tion.
8 I Other Applications of the Hermite Normal Form
The Hermite normal form and the elementary row operations provide
techniques for dealing with problems we have already encountered and
handled rather awkwardly.
A Standard Basis for a Subspace
Let A = {a l5 . . . , aj be a basis of U and let W be a subspace of U spanned
by the set B = {/? l5 . . . , /?,.}. Since every subspace of U is spanned by a finite
set, it is no restriction to assume that B is finite. Let & = 2"=i ^a^i s0 tnat
(b a , . . . , b in ) is the «tuple representing fa. Then in the matrix B = [b^]
each row is the representation of a vector in B. Now suppose an elementary
row operation is applied to B to obtain B'. Every row of B' is a linear com
bination of the rows of B and, since an elementary row operation has an
inverse, every row of B is a linear combination of the rows of B' . Thus the
rows of B and the rows of B' represent sets spanning the same subspace W.
We can therefore reduce B to Hermite normal form and obtain a particular
set spanning W. Since the nonzero rows of the Hermite normal form are
linearly independent, they form a basis of W.
8  Other Applications of the Hermite Normal Form 69
Now let C be another set spanning W. In a similar fashion we can con
struct a matrix C whose rows represent the vectors in C and reduce this
matrix to Hermite normal form. Let C" be the Hermite normal form
obtained from C, and let B' be the Hermite normal form obtained from B.
We do not assume that B and C have the same number of elements, and there
fore B' and C do not necessarily have the same number of rows. However,
in each the number of nonzero rows must be equal to the dimension of W.
We claim that the nonzero rows in these two normal forms are identical.
To see this, construct a new matrix with the nonzero rows of C" written
beneath the nonzero rows of B' and reduce this matrix to Hermite normal
form. Since the rows of C are dependent on the rows of B', the rows of C"
can be removed by elementary operations, leaving the rows of B'. Further
reduction is not possible since B' is already in normal form. But by inter
changing rows, which are elementary operations, we can obtain a matrix in
which the nonzero rows of B' are beneath the nonzero rows of C". As
before, we can remove the rows of B' leaving the nonzero rows of C as
the normal form. Since the Hermite normal form is unique, we see that the
nonzero rows of B' and C are identical. The basis that we obtain from the
nonzero rows of the Hermite normal form is the standard basis with respect
to A for the subspace W.
This gives us an effective method for deciding when two sets span the
same subspace. For example, in Chapter 14, Exercise 5, we were asked to
show that {(1, 1,0, 0), (1, 0, 1, 1)} and {(2, 1, 3, 3), (0, 1, 1, 1)} span
the same space. In either case we obtain {(1, 0, 1, 1), (0, 1, —1, —1)} as the
standard basis.
The Sum of Two Subspaces
If A x is a subset spanning W x and A 2 is a subset spanning VV 2 , then A x U A 2
spans W 1 + W 2 (Chapter I, Proposition 4.4). Thus we can find a basis for
W x + VV 2 by constructing a large matrix whose rows are the representations
of the vectors in A x u A 2 and reducing it to Hermite normal form by ele
mentary row operations.
The Characterization of a Subspace by a Set of Homogeneous
Linear Equations
We have already seen that the set of all solutions of a system of homo
geneous linear equations is a subspace, the kernel of the linear transformation
represented by the matrix of coefficients. The method for solving such a
system which we described in Section 7 amounts to passing from a charac
terization of a subspace as the set of all solutions of a system of equations
to its description as the set of all linear combinations of a basis. The question
70 Linear Transformations and Matrices  II
naturally arises: If we are given a spanning set for a subspace W, how can
we find a system of simultaneous homogeneous linear equations for which W
is exactly the set of solutions ?
This is not at all difficult and no new procedures are required. All that
is needed is a new look at what we have already done. Consider the homo
geneous linear equation a x x x \ + a n x n = 0. There is no significant
difference between the a/s and the s/s in this equation ; they appear sym
metrically. Let us exploit this symmetry systematically.
If a 1 x 1 + • • • + a n x n = and b x x x + \ b n x n = are two homo
geneous linear equations then (a x + b x )x x + • • • + (a n + b n )x n = is a
homogeneous linear equation as also is aa 1 x x + • • • + aa n x n = where
a e F. Thus we can consider the set of all homogeneous linear equations in
n unknowns as a vector space over F. The equation a x x x + • • • + a n x n =
is represented by the «tuple (a x , . . . , a n ).
When we write a matrix to represent a system of equations and reduce that
matrix to Hermite normal form we are finding a standard basis for the sub
space of the vector space of all homogeneous linear equations in x x , . . . , x n
spanned by this system of equations just as we did in the first part of this
section for a set of vectors spanning a subspace. The rank of the system of
equations is the dimension of the subspace of equations spanned by the given
system.
Now let W be a subspace given by a spanning set and solve for the subspace
£ of all equations satisfied by W. Then solve for the subspace of solutions of
the system of equations £. W must be a subspace of the set of all solutions.
Let W be of dimension v. By Theorem 7.1 the dimension of £ is n — v.
Then, in turn, the dimension of the set of all solutions of £ is n — (n — v) = v.
Thus W must be exactly the space of all solutions. Thus W and £ characterize
each other.
If we start with a system of equations and solve it by means of the Hermite
normal form, as described in Section 7, we obtain in a natural way a basis
for the subspace of solutions. This basis, however, will not be the standard
basis. We can obtain full symmetry between the standard system of equations
and the standard basis by changing the definition of the standard basis.
Instead of applying the elementary row operations by starting with the left
hand column, start with the righthand column. If the basis obtained in this
way is called the standard basis, the equations obtained will be the standard
equations, and the solution of the standard equations will be the standard
basis. In the following example the computations will be carried out in this
way to illustrate this idea. It is not recommended, however, that this be
generally done since accuracy with one definite routine is more important.
Let
W = <(1, 0, 3, 11, 5), (3, 2, 5, 5, 3), (1, 1,2, 4, 2), (7, 2, 12, 1, 2)>.
8  Other Applications of the Hermite Normal Form
We now find a standard basis by reducing
71
3
2 5
to the form
2
12
5
2
11
5
4
1
1
1
5
3
2
2
From this we see that the coefficients of our systems of equations satisfy the
conditions
2a x + 5a z + a 5 =
a x + 2a z + a 4 =0
a x + a 2 = 0.
The coefficients a x and a z can be selected arbitrarily and the others computed
from them. In particular, we have
(a x , a 2 , a 3 , a 4 , a b ) = a x (l, 1, 0, 1, 2) + a 3 (0, 0, 1, 2, 5).
The 5tuples (1,1,0,1,2) and (0,0, 1, 2, 5) represent the two
standard linear equations
x l ~ x 2 ~ x i — 2x 5 =
The reader should check that the vectors in W actually satisfy these equations
and that the standard basis for W is obtained.
The Intersection of Two Subspaces
Let W x and W 2 be subspaces of U of dimensions v x and i> 2 , respectively,
and let W x n VV 2 be of dimension v. Then W x + VV 2 is of dimension
v i + v 2 — v. Let E x and £ 2 be the spaces of equations characterizing W x
and W 2 . As we have seen E x is of dimension n — v x and £ 2 is of dimension
n — v 2 . Let the dimension of E x + £ 2 be />. Then E x n £ 2 is of dimension
(«  v x ) + (n  v 2 )  p = 2n  v x  v 2  p.
Since the vectors in W x n W 2 satisfy the equations in both £ x and £ 2 ,
they satisfy the equations in E x + £ 2 . Thus v < n — p. On the other hand,
72 Linear Transformations and Matrices  II
W x and VV 2 both satisfy the equations in E x n £ 2 so that W 1 + VV 2 satisfies the
equations in £ x n £ 2 . Thus v x + r 2 — v < n — {2n — v x — v 2 — p) =
v i + ^2 + /° — «• A comparison of these two inequalities shows that
v — n — p and hence that W x n VV 2 is characterized by £ x + £ 2 .
Given W x and W 2 , the easiest way to find W x n W 2 is to determine £ x and
£ 2 and then £ x + £ 2  From £ x + £ 2 we can then find W x n VV 2 . In effect,
this involves solving three systems of equations, and reducing to Hermit
normal form three times, but it is still easier than a direct assault on the
problem.
As an example consider Exercise 8 of Chapter 14. Let W x = ((1,2, 3, 6),
(4,1,3,6), (5, 1,6, 12)) and W 2 = ((1,1,1,1), (2,1,4,5)). Using
the Hermite normal form, we find that £ x = <(— 2, — 2, 0, 1), (— 1, — 1, 1,0))
and £ 2 = <(4, 3,0, 1), (3,2,1,0)). Again, using the Hermite
normal form we find that the standard basis for Ej + £ 2 is {(1, 0, 0, £),
(0,1,0,1), (0,0,1,— )}. And from this we find quite easily that,
W,nW 2 = ((i,l, i, 1)>.
Let B = {/?!, ^ 2 , . . . , (5 n ) be a given finite set of vectors. We wish to
solve the problem posed in Theorem 2.2 of Chapter I. How do we show that
some fi k is a linear combination of the & with i < k; or how do we show that
no ft k can be so represented ?
We are looking for a relation of the form
& i
P* = I ***&• « (8.1)
This is not a meaningful numerical problem unless ^ is a given specific set.
This usually means that the & are given in terms of some coordinate system,
relative to some given basis. But the relation (8.1) is independent of any
coordinate system so we are free to choose a different coordinate system if this
will make the solution any easier. It turns out that the tools to solve this
problem are available.
Let A = {<*!, . . . , a TO } be the given basis and let
m
Pi = Z a » a *' ;' = 1, . . . , n. (8.2)
If A' = {a{, . . . , a' m } is the new basis (which we have not specified yet),
we would have
m
Pi = 2 a 'n^ j =1,. . . ,n. (8.3)
i=l
What is the relation between A = [a tj ] and A' = [a' i} ]7 If P is the matrix of
transition from the basis A to the basis A', by formula (4.3) we see that
A = PA'. (8.4)
8  Other Applications of the Hermite Normal Form 73
Since P is nonsingular, it can be represented as a product of elementary
matrices. This means A' can be obtained from A by a sequence of elementary
row operations.
The solution to (8.1) is now most conveniently obtained if we take A' to be
in Hermite normal form. Suppose that A' is in Hermite normal form and
use the notation given in Theorem 5.1. Then, for P k . we would have
Pki = <4 (8.5)
and for j between k r and k r+1 we would have
r
Pi = 2 a i><
= 2 a 'iAi
(8.6)
Since k t < k r < j, this last expression is a relation of the required form.
(Actually, every linear relation that exists among the j8 < can be obtained
from those in (8.6). This assertion will not be used later in the book so we
will not take space to prove it. Consider it "an exercise for the reader.")
Since the columns of A and A' represent the vectors in 8, the rank of A is
equal to the number of vectors in a maximal linearly independent subset of 8.
Thus, if 8 is linearly independent the rank of A will be n, this means that the
Hermite normal form of A will either show that 6 is linearly independent
or reveal a linear relation in 8 if it is dependent.
For example, consider the set {(1,0,3,11,5), (3,2,5,5, 3),
(1, 1,2, —4, 2), (7, 2, 12, 1, 2)}. The implied context is that a basis A =
{a l5 . . . , a 5 } is considered to be given and that & = a x — 3a 3 + lla 4 — 5a 5
etc. According to (8.2) the appropriate matrix is
"13 1 T
11 5 4 1
5 3 2 2_
which reduces to the Hermite normal form
3
2
12
"1
3"
4
1
2 3
4"
1
a 9
2"
74 Linear Transformations and Matrices  II
It is easily checked that — 1(1, 0, 3, 11, 5) +  2 T 3 (3, 2, 5, 5, 3) 
V 9 0,l,2, 4, 2) = (7, 2, 12, 1,2).
EXERCISES
1 . Determine which of the following set in R 4 are linearly independent over R.
(a) {(1,1,0,1), (1, 1,1,1), (2,2, 1,2), (0,1,0,0)}.
(b) {(1, 0, 0, 1), (0, 1, 1, 0), (1, 0, 1, 0), (0, 1, 0, 1)}.
(c) {(1, 0, 0, 1), (0, 1, 0, 1), (0, 0, 1, 1), (1, 1, 1, 1)}.
This problem is identical to Exercise 8, Chapter 12.
2. Let W be the subspace of R 5 spanned by {(1,1,1,1,1), (1,0,1,0,1),
(0,1,1,1,0), (2,0,0,1,1), (2,1,1,2,1), (1, 1, 1, 2,2), (1,2,3,4, 1)}.
Find a standard basis for W and the dimension of W. This problem is identical
to Exercise 6, Chapter 14.
3. Show that {(1, 1,2, 3), (1,1,2,0), (3, 1,6, 6)} and {(1,0,1,0),
(0, 2, 0, 3)} do not span the same subspace. This problem is identical to Exercise 7,
Chapter 14.
4. If W x = <(1, 1, 3, 1), (1, 0, 2, 0), (3, 2, 4, 2)> and W 2 = <(1, 0, 0, 1),
(1, 1,7, 1)> determine the dimension of W x + W 2 .
5. Let W = <(1, 1, 3, 0, 1), (2, 1,0, 1,4), (3, 1,1,1, 8), (1, 2, 3, 2, 6)>.
Determine the standard basis for W. Find a set of linear equations which char
acterize W.
6. Let W x = <(1, 2, 3, 6), (4, 1, 3, 6), (5, 1, 6, 12)> and W 2 = <(1, 1, 1, 1),
(2,  1 , 4, 5)> be subspaces of R 4 . Find bases for W x n W 2 and W x + VV 2 . Extend
the basis of VV 2 n W 2 to a basis of W x and extend the basis of W 1 n W 2 to a basis
of VV 2 . From these bases obtain a basis of W x + W 2 . This problem is identical
to Exercise 8, Chapter 14.
9 I Normal Forms
To understand fully what a normal form is, we must first introduce the
concept of an equivalence relation. We say that a relation is defined in a
set if, for each pair {a, b) of elements in this set, it is decided that "a is
related to Z>" or "a is not related to 6." If a is related to b, we write a~ b.
An equivalence relation in a set S is a relation in S satisfying the following laws :
Reflexive law: a~ a,
Symmetric law: If a <~ b, then b <~ a.
Transitive law : If a ~ b and b ~ c, then a ^ c.
If for an equivalence relation we have a ~ b, we say that a is equivalent to b.
9  Normal Forms 75
Examples. Among rational fractions we can define a\b ~ c\d (for a, b,c,d
integers) if and only if ad = be. This is the ordinary definition of equality
in rational numbers, and this relation satisfies the three conditions of an
equivalence relation.
In geometry we do not ordinarily say that a straight line is parallel to
itself. But if we agree to say that a straight line is parallel to itself, the
concept of parallelism is an equivalence relation among the straight lines
in the plane or in space.
Geometry has many equivalence relations: congruence of triangles,
similarity of triangles, the concept of projectivity in projective geometry,
etc. In dealing with time we use many equivalence relations: same hour
of the day, same day of the week, etc. An equivalence relation is like a
generalized equality. Elements which are equivalent share some common
or underlying property. As an example of this idea, consider a collection of
sets. We say that two sets are equivalent if their elements can be put into
a onetoone correspondence; for example, a set of three battleships and a
set of three cigars are equivalent. Any set of three objects shares with
any other set of three objects a concept which we have abstracted and called
"three." All other qualities which these sets may have are ignored.
It is most natural, therefore, to group mutually equivalent elements
together into classes which we call equivalence classes. Let us be specific
about how this is done. For each a e S, let S a be the set of all elements in
S equivalent to a; that is, b e S a if and only if b ~ a. We wish to show that
the various sets we have thus defined are either disjoint or identical.
Suppose S n S b is not empty; that is, there exists a cES a nS b such
that c r^j a and c ~ b. By symmetry b <~ c, and by transitivity b — a.
If d is any element of S 6 , d^b and hence d~ a. Thus d eS a and S 6 <= S a .
Since the relation between S and S b is symmetric we also have S <=■ S b and
hence S a = S b . Since a eS u we have shown, in effect, that a proposed
equivalence class can be identified by any element in it. An element selected
from an equivalence class will be called a representative of that class.
An equivalence relation in a set S defines a partition of that set into
equivalence classes in the following sense: (1) Every element of S is in some
equivalence class, namely, a e S . (2) Two elements are in the same equiva
lence class if and only if they are equivalent. (3) Nonidentical equivalence
classes are disjoint. On the other hand, a partition of a set into disjoint
subsets can be used to define an equivalence relation; two elements are
equivalent if and only if they are in the same subset.
The notions of equivalence relations and equivalence classes are not
nearly so novel as they may seem at first. Most students have encountered
these ideas before, although sometimes in hidden forms. For example,
we may say that two differentiable functions are equivalent if and only if
76 Linear Transformations and Matrices  II
they have the same derivative. In calculus we use the letter "C" in describing
the equivalence classes; for example, x 3 + # 2 + 2z + C is the set (equiva
lence class) of all functions whose derivative is 3x 2 + 2x + 2.
In our study of matrices we have so far encountered four different equiva
lence relations:
I. The matrices A and B are said to be left associate if there exists a
nonsingular matrix Q such that B = Q~ X A. Multiplication by Q~ x corre
sponds to performing a sequence of elementary row operations. If A
represents a linear transformation a of U into V with respect to a basis A in U
and a basis B in V, the matrix B represents a with respect to A and a new basis
in V.
II. The matrices A and B are said to be right associate if there exists a
nonsingular matrix P such that B = AP.
III. The matrices A and B are said to be associate if there exist non
singular matrixes P and Q such that B = Q~ X AP. The term "associate" is
not a standard term for this equivalence relation, the term most frequently
used being "equivalent." It seems unnecessarily confusing to use the same
term for one particular relation and for a whole class of relations. Moreover,
this equivalence relation is perhaps the least interesting of the equivalence
relations we shall study.
IV. The matrices A and B are said to be similar if there exists a non
singular matrix P such that B = P~ X AP. As we have seen (Section 4) similar
matrices are representations of a single linear transformation of a vector
space into itself. This is one of the most interesting of the equivalence
relations, and Chapter III is devoted to a study of it.
Let us show in detail that the reation we have defined as left associate
is an equivalence relation. The matrix Q~ x appears in the definition because
Q represents the matrix of transition. However, Qr x is just another
singular matrix, so it is clearly the same thing to say that A and B are left
associate if and only if there exists a nonsingular matrix Q such that B = QA.
(1) A — ' A since IA = A.
(2) If A ~ B, there is a nonsingular matrix Q such that B = QA. But
then A = Q~ X B so that B ~ A.
(3) If A r^j B and B ~ C, there exist nonsingular matrices Q and P
such that B = QA and C = PB. But then PQA = PB = C and PQ is
nonsingular so that A ~ C.
For a given type of equivalence relation among matrices a normal form
is a particular matrix chosen from each equivalence class. It is a repre
sentative of the entire class of equivalent matrices. In mathematics the
terms "normal" and "canonical" are frequently used to mean "standard"
in some particular sense. A normal form or canonical form is a standard
9  Normal Forms 77
form selected to represent a class of equivalent elements. A normal form
should be selected to have the following two properties : Given any matrix
A, (1) it should be possible by fairly direct and convenient methods to find
the normal form of the equivalence class containing A, and (2) the method
should lead to a unique normal form.
Often the definition of a normal form is compromised with respect to the
second of these desirable properties. For example, if the normal form were
a matrix with complex numbers in the main diagonal and zeros elsewhere,
to make the normal form unique it would be necessary to specify the order
of the numbers in the main diagonal. But it is usually sufficient to know
the numbers in the main diagonal without regard to their order, so it would
be an awkward complication to have to specify their order.
Normal forms have several uses. Perhaps the most important use is that
the normal form should yield important or useful information about the
concept that the matrix represents. This should be amply illustrated in
the case of the concept of left associate and the Hermite normal form. We
introduced the Hermite normal form through linear transformations, but
we found that it yielded very useful information when the matrix was used
to represent linear equations or bases of subspaces.
Given two matrices, we can use the normal form to tell whether they are
equivalent. It is often easier to reduce each to normal form and compare
the normal forms than it is to transform one into the other. This is the case,
for example, in the application described in the first part of Section 8.
Sometimes, knowing the general appearance of the normal form, we can
find all the information we need without actually obtaining the normal
form. This is the case for the equivalence relation we have* called associate.
The normal form for this equivalence relation is described in Theorem 4.1.
T*^re is just one normal form for each possible value of the rank. The
lia'mber of different equivalence classes is min {m, n} + 1. With this notion
of equivalence the rank of a matrix is the only property of importance. Any
two matrices of the same rank are associate. In practice we can find the rank
without actually computing the normal form of Theorem 4.1. And knowing
the rank we know the normal form.
We encounter several more equivalence relations among matrices. The
type of equivalence introduced will depend entirely on the underlying con
cepts the matrices are used to represent. It is worth mentioning that for the
equivalence relations we introduce there is no necessity to prove, as we did
for an example above, that each is an equivalence relation. An underlying
concept will be defined without reference to any coordinate system or choice
of basis. The matrices representing this concept will transform according
to certain rules when the basis is changed. Since a given basis can be retained
the relation defined is reflexive. Since a basis changed can be changed back
For example, in R 2 the matrix
1 0"
^axis and
78 Linear Transformations and Matrices  II
to the original basis, the relation defined is symmetric. A basis changed once
and then changed again depends only on the final choice so that the relation is
transitive.
For a fixed basis A in U and 8 in V two different linear transformations
a and t of U into V are represented by different matrices. If it is possible,
however, to choose bases A' in U and 8' in V such that the matrix representing
r with respect to A' and 8' is the same as the matrix representing a with
respect to A and 8, then it is certainly clear that a and t share important
geometric properties.
For a fixed a two matrices A and A' representing a with respect to different
bases are related by a matrix equation of the form A' = Q~ X AP. Since
A and A' represent the same linear transformation we feel that they should
have some properties in common, those dependent upon a.
These two points of view are really slightly different views of the same
kind of relationship. In the second case, we can consider A and A' as
representing two linear transformations with respect to the same basis,
instead of the same linear transformation with respect to different bases.
"1 01
represents a reflection about the
lj
represents a reflection about the # 2 axis. When both
1.
linear transformations are referred to the same coordinate system they are
different. However, for the purpose of discussing properties independent
of a coordinate system they are essentially alike. The study of equivalence
relations is motivated by such considerations, and the study of normal forms
is aimed at determining just what these common properties are that are
shared by equivalent linear transformations or equivalent matrices.
To make these ideas precise, let a and r be linear transformations of V
into itself. We say that a and t are similar if there exist bases A and 8 of V
such that the matrix representing a with respect to A is the same as the matrix
representing t with respect to 8. If A and B are the matrices representing
a and t with respect to A and P is the matrix of transition from A to 8, then
P~ X BP is the matrix representing r with respect to 8. Thus a and r are
similar if P^BP = A.
In a similar way we can define the concepts of left associate, right associate,
and associate for linear transformations.
*10 I Quotient Sets, Quotient Spaces
Definition. If S is any set on which an equivalence relation is defined, the
collection of equivalence classes is called the quotient or factor set. Let S
denote the quotient set. An element of S is an equivalence class. If a is an
10 I Quotient Sets, Quotient Spaces 79
element of S and a is the equivalence class containing a, the mapping t] that
maps a onto a is well defined. This mapping is called the canonical mapping.
Although the concept of a quotient set might appear new to some, it is
certain that almost everyone has encountered the idea before, perhaps in
one guise or another. One example occurs in arithmetic. In this setting,
let S be the set of all formal fractions of the form afb where a and b are
integers and b ^ 0. Two such fractions, ajb and c/d, are equivalent if and
only if ad = be. Each equivalence class corresponds to a single rational
number. The rules of arithmetic provide methods of computing with rational
numbers by performing appropriate operations with formal fractions selected
from the corresponding equivalence classes.
Let U be a vector space over F and let K be a subspace of U. We shall call
two vectors a, p e U equivalent modulo K if and only if their difference lies
in K. Thus a /^ < p if and only if a — p e K. We must first show this defines
an equivalence relation. (1) a ~ a because a — a = e K. (2)a^/?=>
a — p 6 K => p — a e K => p ~ a. (3) {a ~ p and p ~ y} => {a  e K and
p — y eK}. Since K is a subspace a — 7 = (a — /?) + (0 — y) e K and,
hence, a ~ y. Thus "~" is an equivalence relation.
We wish to define vector addition and scalar multiplication in U. For
a e U, let SeU denote the equivalence class containing a. a is called a
representative of a. Since a may contain other elements besides a, it may
happen that a^a' and yet a = a'. Let a and p be two elements in U. Since
a, /? e U, a + jS is defined. We wish to define a + /? to be the sum of a and /5.
In order for this to be well defined we must end up with the same equivalence
class as the sum if different representatives are chosen from a and /?. Suppose
a = a' a nd,g = /?' . Then a  a' £ K,  ? e K, and (a + $)  (a' + ?) £
/C. Thus a + = a' + /?' and the sum is well defined. Scalar multiplication
is defined similarly. For aeF, aa. is thus defined to be the equivalence class
containing ace; that is, aa = aa These operations in (Jare said to be induced
by the corresponding operation in U.
Theorem 10.1. If U is a vector space over F, and K is a subspace of U,
the quotient set U with vector addition and scalar multiplication defined as
above is a vector space over F.
proof. We leave this as an exercise. □
For any a £ U, the symbol a + K is used to denote the set of all elements
in U that can be written in the form a + y where y £ K. (Strictly speaking,
we should denote the set by {a} + K so that the plus sign combines two objects
of the same type. The notation introduced here is traditional and simpler.)
The set a + K is called a coset of K. If /? £ a + K, then p — a £ K and
80 Linear Transformations and Matrices  II
j8 ~ a. Conversely, if ~ a, then £  a = y e K so /S e a + K. Thus
a + K is simply the equivalence class a containing a. Thus a + K = /? + K
if and only ifae£ = £ + Kor£ea = a + K.
The notation a + K to denote a is convenient to use in some calcula tions.
For example, a + = (a + K) + (0 + K) = a + + K = a + 0, and
aa = a(a + K) = aa + aK c fla + K = tfa. Notice that aa. = act. when a
and ool are considered to be elements of U and scalar mutliplication is the
induced operation, but that ail and ad. may not be the same when they are
viewed as subsets of U (for example, let a = 0). However, since acx c: ace
the set acl determines the desired coset in U for the induced operations. Thus
we can compute effectively in U by doing the corresponding operations with
representatives. This is precisely what is done when we compute in residue
classes of integers modulo an integer m.
Definition. U with the induced operations is called a factor space or quotient
space. In order to designate the role of the subspace K which defines the
equivalence relations, U is usually denoted by U/K.
In our discussion of solutions of linear problems we actually encountered
quotient spaces, but the discussion was worded in such a way as to avoid
introducing this more sophisticated concept. Given the linear transformation
a of U into V, let K be the kernel of a and let U = U/K be the corresponding
quotient space. If a x and a 2 are solutions of the linear problem, oc(f) = 0,
then a(a. x — <x 2 ) = so that a x and a 2 are in the same coset of K. Thus
for each fi e lm(a) there corresponds precisely one coset of K. In fact
the correspondence between U/K and lm(a) is an isomorphism, a fact which
is made more precise in the following theorem.
Theorem 10.2. (First homomorphism theorem). Let a be a linear trans
formation ofU into V. Let K be the kernel of a. Then a can be written as the
product of a canonical mapping rj of U onto = U/K and a monomorphism
a x of U into V.
proof. The canonical mapping r\ has already been defined. To define
a lf for each a e U let (7 x (a) = cr(a) where a is any representative of a. Since
(7(a) = (r(a') for a~a', a x is well defined. It is easily seen that a x is a
monomorphism since a must have different values in different cosets. D
The homomorphism theorem is usually stated by saying, "The homo
morphic image is isomorphic to the quotient space of U modulo the kernel."
Theorem 10.3. (Mapping decomposition theorem). Let a be a linear
transformation ofil into V. Let K be the kernel of a and I the image of a. Then
a can be written as the product a = ia x rj, where rj is the canonical mapping of
10 I Quotient Sets, Quotient Spaces 81
U onto U = UjK, a 1 is an isomorphism of U onto I, and i is the injection of \
into V.
proof. Let a' be the linear transformation of U onto / induced by re
stricting the codomain of a to the image of a. By Theorem 10.2, a' can be
written in the form a' = a^. □
Theorem 10.4. {Mapping factor theorem). Let S be a subspace of U and
let U = U/S be the resulting quotient space. Let a be a linear transformation of
U into V, and let K be the kernel of a. If S <= K, then there exists a linear
transformation a x of U into V such that a = a x r\ where n is the canonical
mapping of U onto U.
proof. For each oteU, let cr^a) = c(a) where a e a. If a' is another
representative of a, then a — a' £ S <= K. Thus <r(a) = cT(a') and a x is well
defined. It is easy to check that g x is linear. Clearly, <r(<x) = o^a) = ^(^(a))
for all a e U, and a = a x rj. □
We say that a factors through (J.
Note that the homomorphism theorem is a special case of the factor theorem
in which K = S.
Theorem 10.5. {Induced mapping theorem). Let U and V be vector spaces
over F, and let r be a linear transformation of (J into V. Let U be a subspace of
U and let V be a subspace of V. If t{U ) <= V , // is possible to define in a
natural way a mapping f of U/U into V/V such that o 2 t = fa x where a x is the
canonical mapping U onto U and a 2 is the canonical mapping of V onto V.
proof. Consider a = a 2 r, which maps U into V. The kernel of a is t 1 (V ).
By assumption, U <= r _1 (V ). Hence, by the mapping factor theorem, there
is a linear transformation f such that fa x = a 2 r. □
We say that f is induced by r.
Numerical calculations with quotient spaces can usually be avoided in
problems involving finite dimensional vector spaces. If U is a vector space
over F and K is a subspace of U, we know from Theorem 4.9 of Chapter I
that K is a direct summand. Let U = K ® W. Then the canonical mapping
r\ maps W isomorphically onto U/K. Thus any calculation involving UjK
can be carried out in W.
Although there are many possible choices for the complementary subspace
W, the Hermite normal form provides a simple and effective way to select a W
and a basis for it. This typically arises in connection with a linear problem.
To see this, reexamine the proof of Theorem 5.1. There we let k ± , k 2 , . . . , k p
be those indices for which <r(a fc .) ^ a{U kil ). We showed there that {(![, . . . ,
fi' k } where ^ = cr(a fc .) formed a basis of a{U). {a &i , a^, . . . , <x k J is a basis for
a suitable W which is complementary to K{a).
82
Linear Transformations and Matrices I II
Example. Consider the linear transformation a of R 5 into R 3 represented by
the matrix
1 1 1"
It is easy to determine that the kernel K of a is 2dimensional with basis
{(1, —1, — 1, 1,0), (0, 0, —1,0, 1)}. This means that a has rank 3 and the
image of a is all of R 3 . Thus R 5 = R 5 /K is isomorphic to R 3 .
Consider the problem of solving the equation c(£) = /?, where jS is
represented by (b x , b 2 , b 3 ). To solve this problem we reduce the augmented
matrix
"1 1 1 b x
1 1 b 2
1 01 b 3
to the Hermite normal form
"1 1 b 3
1 1 b 2
1 1 1 b x  I
K
This means the solution £ is represented by
(b 3 ,b 2 ,b x b 3 ,0,0) + x i (l, l,l,l,0) + z 5 (0, 0,1,0,1).
(ft,, b t , b x  b 3 , 0, 0) = M0, 0,1,0,0) + 6,(0, 1,0,0,0) + b 3 (\ ,0,1,0,0)
is a particular solution and a convenient basis for a subspace W complementary
to K is {(0, 0, 1,0, 0), (0, 1, 0, 0, 0), (1, 0, 1, 0, 0)}. a maps Z> x (0, 0,
1, 0, 0) + b 2 (0, 1, 0, 0, 0) + b 3 (l, 0, 1, 0, 0) onto (b lt b 2 , b 3 ). Hence, W
is mapped isomorphically onto R 3 .
This example also provides an opportunity to illustrate the working of the
first homomorphism theorem. For any (x x , x 2 , x 3 , x A , x 5 ) e R 5 .
 i
(*!, X 2 , X 3 , Xi Hr X b ) = (X X + X 3 + Xs)(0, 0, 1 , 0, 0)
+ (*2 + *4)(0, 1,0,0,0)'
+ (^^(1, 0,1,0,0)
+ * 4 (1, 1, l,l,0) + a; 5 (0, 0,1,0,1).
Thus {x x , x 2 , x 3 , x 4 , x 5 ) is mapped onto the coset (x x + x 3 + x 5 )(0, 0, 1, 0, 0) +
(x 2 + a; 4 )(0, 1, 0, 0, 0) + (x x  a; 4 )(l, 0, 1, 0, 0) + K under the natural
homomorphism onto R 5 /K. This coset is then mapped isomorphically onto
{x x + x 3 + x h , x 2 + x x , x x — x y ) e R 3 . However, it is somewhat contrived to
llHom(U,V) 83
work out an example of this type. The main importance of the first homo
morphism theorem is theoretical and not computational.
*11 I Hom(U, V)
Let U and V be vector spaces over F. We have already observed in Section 1
that the set of all linear transformations of U into Vcan be made into a vector
space over F by defining addition and scalar multiplication appropriately.
In this section we will explore some of the elementary consequences of this
observation. We shall call this vector space Hom(U, V), "The space of all
homomorphisms of U into V."
Theorem 11.1. If dim U = n and dim V = m, then dim Hom(U, V) = mn.
proof. Let {a x , . . . , a„} be a basis of U and let {&, . . . , fi n } be a basis of
V. Define the linear transformation of a tj by the rule
°wO*) = <*«&
TO
= ZWr (n.i)
r=l
Thus a it is represented by the matrix [d ri d ik ] = A u . A it has a zero in every
position except for a 1 in row / column j.
The set {a u } is linearly independent. For if a linear relation existed among
the o u it would be of the form
1 au^ii = 0.
This means 2. .fl 4 ,(T, y (a fc ) = for alia*. But J., ^(T,,(a ft ) = 2 w a w <5 y *ft =
Zi a ikPi = 0 Since {&} is a lineary independent set, a ik = for / = 1 ,
2, . . . , m. Since this is true for each k, all a u = and {a it } is linearly
independent.
If or e Hom(U, V) and cr(a fc ) = ^tt a*Pi,
then
m / n \
m n
= 2 2>«o«(a*)
i=l 3=1 J
Thus {or„} spans Hom(U, V), which is therefore of dimension mn. a
If V x is a subspace of V, every linear transformation of U into V x also defines
a mapping of U into V. This mapping of U into V is a linear transformation of
84 Linear Transformations and Matrices  II
U into V. Thus, with each element of Hom(U, Vj) there is associated in a
natural way an element of Hom(U, V). We can identify Hom(U, V x ) with
asubsetofHom(U, V). With this identification Horn (U, V x ) is a subspace of
Hom(U, V).
Now let U 1 be a subspace of U. In this case we cannot consider Hom(U x , V)
to be a subset of Hom(U, V) since a linear transformation in Hom(L/ 1 , V)
is not necessarily defined on all of U. But any linear transformation in
Hom(U, V) is certainly defined on U x . If a e Hom(U, V) we shall consider
the mapping obtained by applying a only to elements in U x to be a new function
and denote it by R(a). R(a) is called the restriction of a to U x . We can con
sider R(a) to be an element of Hom(L/ x , V).
It may happen that different linear transformations defined on U produce
the same restriction on U x . We say that o x and a 2 are equivalent on L/ x if and
only if R(a x ) = R(a 2 ). It is clear that R(a + r) = i*(<r) + i?(r) and R(aa) =
aR(a) so that the mapping of Hom(U, V) into Kom(U x , V) is linear. We call
this mapping R, the restriction mapping.
The kernel of R is clearly the set of all linear transformations in Hom(U, V)
that vanish on U x . Let us denote this kernel by U*.
If a is any linear transformation belonging to Hom(U x , V), it can be
extended to a linear transformation belonging to Hom(U, V) in many ways.
If {a x , . . . , <xj is a basis of U such that {a l5 . . . , a r } is a basis of U x , then let
(7(a,) = <r(a 3 ) for j = 1 , . . . , r, and let crfo) be defined arbitrarily for
j = r + 1 , . . . , n. Since a is then the restriction of a, we see that R is an
epimorphism of Hom(U, V) onto Hom^, V). Since Hom(U, V) is of dimen
sion mn and Hom(U x , V) is of dimension mr, U* is of dimension m(n — r).
Theorem 11.2. Hom(L/ x , V) is canonically isomorphic to Hom(U, V)/U*. □
Note: It helps the intuitive understanding of this theorem to examine the
method by which we obtained an extension off on U x , to a on U. U* is the
set of all extensions of a when a is the zero mapping, and one can see directly
that the dimension of U* is (n — r)m.
chapter
III
Determinants,
eigenvalues,
and similarity
transforma
tions
This chapter is devoted to the study of matrices representing linear trans
formations of a vector space into itself. We have seen that if A represents
a linear transformation a of V into itself with respect to a basis A, and P
is the matrix of transition from A to a new basis A', then P~ X AP = A' is
the matrix representing a with respect to A'. In this case A and A' are said
to be similar and the mapping of A onto A' = P~ X AP is called a similarity
transformation (on the set of matrices, not on V).
Given a, we seek a basis for which the matrix representing a is particularly
simple. In practice a is given only implicitly by giving a matrix A representing
a. The problem, then, is to determine the matrix of transition P so that
P~ X AP has the desired form. The matrix representing a has its simplest
form whenever a maps each basis vector onto a multiple of itself; that is,
whenever for each basis vector a there exists a scalar A such that <r(a) = Aa.
It is not always possible to find such a basis, but there are some rather general
conditions under which it is possible. These conditions include most cases
of interest in the applications of this theory to physical problems.
The problem of finding nonzero a such that cr(oc) = Aa is equivalent to
the problem of finding nonzero vectors in the kernel of a — A. This is a
linear problem and we have given practical methods for solving it. But
there is no nonzero solution to this problem unless a — A is singular. Thus
we are faced with the problem of finding those A for which a — X is singular.
The values of X for which a — X is singular are called the eigenvalues of a,
and the nonzero vectors a for which <r(a) = Aa are called eigenvectors of a.
We introduce some topics from the theory of determinants solely for
the purpose of finding the eigenvalues of a linear transformation. Were
it not for this use of determinants we would not discuss them in this book.
Thus, the treatment given them here is very brief.
85
86 Determinants, Eigenvalues, and Similarity Transformations  III
Whenever a basis of eigenvectors exists, the use of determinants will
provide a method for finding the eigenvalues and, knowing the eigenvalues,
use of the Hermite normal form will enable us to find the eigenvectors.
This method is convenient only for vector spaces of relatively small di
mension. For numerical work with large matrices other methods are
required.
The chapter closes with a discussion of what can be done if a basis of
eigenvectors does not exist.
1 I Permutations
To define determinants and handle them we have to know something
about permutations. Accordingly, we introduce permutations in a form
most suitable for our purposes and develop their elementary properties.
A permutation n of a set S is a onetoone mapping of S onto itself. We
are dealing with permutations of finite sets and we take S to be the set of
the first n integers; S == {1, 2, . . . , n). Let n(i) denote the element which
n associates with /. Whenever we wish to specify a particular permutation
we describe it by writing the elements of S in two rows ; the first row con
taining the elements of S in any order and the second row containing the
element n(i) directly below the element i in the first row. Thus for S =
{1, 2, 3, 4}, the permutation n for which n{\) = 2, n(2) = 4, tt(3) = 3,
and 7r(4) = 1 , can conveniently be described by the notations
/I 2 3 4, Qr /2 4 1 3V ^ ,4.32
\2 4 3 1/ \4 1 2 3/ \l 2 3 4
Two permutations acting on the same set of elements can be combined
as functions. Thus, if n and a are two permutations, an will denote that
permutation mapping i onto a[n(i)]; (an)(i) = a[n(i)]. As an example,
let n denote the permutation described above and let
a =
Then
OTT =
12 3 4
13 4 2
12 3 4
3 2 4 1
Notice particularly that an ^ na.
If n and a are two given permutations, there is a unique permutation
p such that pn = a. Since p must satisfy the condition that p[n(i)] = a(i),
p can be described in our notation by writing the elements n(i) in the first
1 I Permutations 87
row and the elements a(i) in the second row. For the n and a described
above,
/2 4 3 1'
13 4 2/.
The permutation that leaves all elements of S fixed is called the identity
permutation and will be denoted by e. For a given tt the unique permutation
7T _1 such that tt~ x tt = e is called the inverse of it.
If for a pair of elements i <y in S we have tt{i) > n(j), we say that tt
performs an inversion. Let k(Tr) denote the total number of inversions
performed by tt; we then say that tt contains k(Tr) inversions. For the
permutation tt described above, k(Tr) = 4. The number of inversions in
7T 1 is equal to the number of inversions in tt.
For a permutation tt, let sgn tt denote the number (— l) kM . "Sgn" is
an abbreviation for "signum" and we use the term "sgn 77" to mean "the
sign of 77." If sgn tt — 1 , we say that tt is even ; if sgn tt = — 1 , we say that
tt is odd.
Theorem 1.1. Sgn air = sgn a • sgn tt.
proof, a can be represented in the form
77(0 • • • TT(j) ■
<y =
OTr(i) • • • Ott(J)
because every element of S appears in the top row. Thus, in counting the
inversions in a it is sufficient to compare 77(1) and tt(j) with OTr(i) and on(j).
For a given i < j there are four possibilities :
1. i <j; tt{i) < 7r(j); ottQ) < ott(j): no inversions.
2. i <j; Tr(i) < tt(j); aTr(i) > ott(j): one inversion in a, one in 077.
3. i <y; 7r(/) > 7r(y); cnr(i) > ott{j)\ one inversion in 77, one in <77r.
4. 1 <y; 7r(/) > 7r(/); (T7r(i) < gtt(j): one inversion in 77, one in a, and
none in cm.
Examination of the above table shows that k{oTr) differs from k(a) + k(ir)
by an even number. Thus sgn an = sgn a • sgn tt. □
Theorem 1.2. If a permutation tt leaves an element of S fixed, the inversions
involving that element need not be considered in determining whether tt is even
or odd.
proof. Suppose 7r(j) = j. There are j — 1 elements of S less than j and
n — j elements of S larger than j. For i < j an inversion occurs if and only
if 7r(0 > tt(j) — j. Let k be the number of elements i in S preceding j for
which 7t(/) > j. Then there must also be exactly k elements i of S following
j for which tt(J) < j. It follows that there are 2k inversions involving j.
Since their number is even they may be ignored in determining sgn tt. □
88 Determinants, Eigenvalues, and Similarity Transformations  III
Theorem 1.3. A permutation which interchanges exactly two elements of
S and leaves all other elements of S fixed is an odd permutation.
proof. Let 77 be a permutation which interchanges the elements / andy
and leaves all other elements of S fixed. According to Theorem 1.2, in
determining sgn tt we can ignore the inversions involving all elements of
S other than i and/ There is just one inversion left to consider and sgn tt =
1. □
Among other things, this shows that there is at least one odd permutation.
In addition, there is at least one even permutation. From this it is but a
step to show that the number of odd permutations is equal to the number of
even permutations.
Let a be a fixed odd permutation. If n is an even permutation, an is odd.
Furthermore, cr _1 is also odd so that to each odd permutation r there cor
responds an even permutation o~ x t. Since a~ 1 {aTr) = tt, the mapping of
the set of even permutations into the set of odd permutations defined by
tt > ott is onetoone and onto. Thus the number of odd permutations is
equal to the number of even permutations.
EXERCISES
1. Show that there are n\ permutations of n objects.
2. There are six permutations of three objects. Determine which of them are
even and which are odd.
3. There are 24 permutations of four objects. By use of Theorem 1.2 and
Exercise 2 we can determine the parity (evenness or oddness) of 15 of these permu
tations without counting inversions. Determine the parity of these 1 5 permutations
by this method and the parity of the remaining nine by any other method.
4. The nine permutations of four objects that leave no object fixed can be
divided into two types of permutations, those that interchange two pairs of objects
and those that permut the four objects in some cyclic order. There are three
permutations of the first type and six of the second. Find them. Knowing the
parity of the 15 permutations that leave at least one object fixed, as in Exercise 3,
and that exactly half of the 24 permutations must be even, determine the parity
of these nine.
5. By counting the inversions determine the parity of
'12 3 4 5
2 4 5 13
Notice that tt permutes the objects in {1, 2, 4} among themselves and the objects
in {3, 5} among themselves. Determine the parity of it on each of these subsets
separately and deduce the parity of tt on all of S.
2 I Determinants
89
2 I Determinants
Let A = [a it ] be a square n x n matrix. We wish to associate with this
matrix a scalar that will in some sense measure the "size" of A and tell us
whether or not A is nonsingular.
Definition. The determinant of the matrix A = [a ti ] is defined to be the
scalar det A = \a it \ computed according to the rule
det A = \a ti \ = 2 ( s 8 n *) a um a **w ' " a
mi(n)>
(2.1)
where the sum is taken over all permutations of the elements of S = {1 , . . . , «}.
Each term of the sum is a product of n elements, each taken from a different
row of A and from a different column of A , and sgn n. The number n is called
the order of the determinant.
As a direct application of this definition we see that
a lx a 12
*2X
= flnfloo — fliofl
12 u 21
fl u fl la tfj
'21
«3i fl a
fl 9
= «U a 22 a 33 + ^12^23^31 + fll3«2i a 32 — «12«21^33
— tf 13 <Z 2 2 fl 31 — Onfl23^32
(2.2)
(2.3)
In general, a determinant of order n will be the sum of n\ products. As
n increases, the amount of computation increases astronomically. Thus it
is very desirable to develop more efficient ways of handling determinants.
Theorem 2.1. det A T = det A.
proof. In the expansion of det A each term is of the form
(Sgn 7r)a lff(1 )a 2ir (2) " ' ' a nrrM
The factors of this term are ordered so that the indices of the rows appear
in the usual order and the column indices appear in a permuted order. In the
expansion of det A T the same factors will appear but they will be ordered
according to the row indices of A T , that is, according to the column indices
of A. Thus this same product will appear in the form
(Sgn 7T 1 )a ff i(i) i ia a i(2),2 ' ' ' «T _1 (n),n
But since sgn tt 1 = sgn tt, this term is identical to the one given above.
Thus, in fact, all the terms in the expansion of det A T are equal to cor
responding terms in the expansion of det A, and det A T = det A. □
90 Determinants, Eigenvalues, and Similarity Transformations  III
A consequence of this discussion is that any property of determinants
developed in terms of the rows (or columns) of A will also imply a cor
responding property in terms of the columns (or rows) of A.
Theorem 2.2. If A' is the matrix obtained from A by multiplying a row
(or column) of A by a scalar c, then det A' = c det A.
proof. Each term of the expansion of det A contains just one element
from each row of A. Thus multiplying a row of A by c introduces the factor c
into each term of det A. Thus det A' = c det A. • □
Theorem 2.3. If A' is the matrix obtained from A by interchanging any
two rows (or columns) of A, then det A' = — det A.
proof. Interchanging two rows of A has the effect of interchanging two
row indices of the elements appearing in A. If or is the permutation inter
changing these two indices, this operation has the effect of replacing each
permutation tt by the permutation tra. Since a is an odd permutation, this
has the effect of changing the sign of every term in the expansion of det A.
Therefore, det A' = —det A. □
Theorem 2.4. If A has two equal rows, det A = 0.
proof. The matrix obtained from A by interchanging the two equal
rows is identical to A, and yet, by Theorem 2.3, this operation must change
the sign of the determinant. Since the only number equal to its negative is
det A = 0. □
Note: There is a minor point to be made here. If 1 + 1 = 0, the proof
of this theorem is not valid, but the theorem is still true. To see this we
return our attention to the definition of a determinant. Sgn tt == 1 for both
even and odd permutations. Then the terms in (2.1) can be grouped into
pairs of equal terms. Since the sum of each pair is 0, the determinant is 0.
Theorem 2.5. If A' is the matrix obtained from A by adding a multiple of
one row (or column) to another, then det A' = det A.
proof. Let A' be the matrix obtained from A by adding c times row k
to rowy. Then
det A' = ^ (sgn 7r)a lw(1) • ■ • (a jwU) + ca kirU) ) • • • a Mk) • • • a nw(n)
IT
= 2 ( s § n ^flird) ' ' ' a trli) • • • a k7r{k) • • • a nn(n)
IT
+ c 2 (sgn 7T)a l1Ta) • • • a k]rU) • • • a kjrik) • • • a nw{n) . (2.4)
ir
The second sum on the right side of this equation is, in effect, the deter
minant of a matrix in which rows j and k are equal. Thus it is zero. The
first term is just the expansion of det A. Therefore, det A' = det A. u
It is evident from the definition that, if / is the identity matrix, det / = 1.
2 I Determinants
91
If E is an elementary matrix of type I, det E = c where c is the scalar
factor employed in the corresponding elementary operation. This follows
from Theorem 2.2 applied to the identity matrix.
If E is an elementary matrix of type II, det E = 1. This follows from
Theorem 2.5 applied to the identity matrix.
If E is an elementary matrix of type III, det E = — 1. This follows from
Theorem 2.3 applied to the identity matrix.
Theorem 2.6. If E is an elementary matrix and A is any matrix, then
det EA = det E • det A = det AE.
proof. This is an immediate consequence of Theorems 2.2, 2.5, 2.3, and
the values of the determinants of the corresponding elementary matrices. □
Theorem 2.7. det A = if and only if A is singular.
proof. If A is nonsingular, it is a product of elementary matrices (see
Chapter II, Theorem 6.1). Repeated application of Theorem 2.6 shows
that det A is equal to the product of the determinants of the corresponding
elementary matrices, and hence is nonzero.
If A is singular, the rows are linearly dependent and one row is a linear
combination of the others. By repeated application of elementary operations
of type II we can obtain a matrix with a row of zeros. The determinant of
this matrix is zero, and by Theorem 2.5 so also is det A. U
Theorem 2.8. If A and B are any two matrices of order n, then det AB =
det A • det B = det BA.
proof. If A and B are nonsingular, the theorem follows by repeated
application of Theorem 2.6. If either matrix is singular, then AB and BA
are also singular and all terms are zero. □
EXERCISES
1. If all elements of a matrix below the main diagonal are zero, the matrix is
said to be in superdiagonal form; that is, a^ = for / > j. If A = [a^] is in super
diagonal form, compute det A.
2. Theorem 2.6 "provides an effective and convenient way to evaluate deter
minants. Verify the following sequence of steps.
3
2
2
1 4
1
1 4
1
1
4
1
= 
3 2
2
= 
10 1
2
4
1
2 4
1
4 1
1 4
1
1 4 1
= 
2
1
= 
2 1
4
1
3
Now use the results of Exercise 1 to evaluate the last determinant.
92
Determinants, Eigenvalues, and Similarity Transformations  III
3. Actually, to compute a determinant there is no need to obtain a superdiagonal
form. And elementary column operations can be used as well as elementary row
operations. Any sequence of steps that will result in a form with a large number of
zero elements will be helpful. Verify the following sequence of steps.
3
2
2
3 2 2
3 2
1
4
1
=
1 4 1
=
1 3 1
2
4
1
10
10
This last determinant can be evaluated by direct use of the definition by computing
just one product. Evaluate this determinant.
4. Evaluate the determinants :
(a)
1 2
1 3
2 5
ib)
1
2
1
1
3
4
1
5
6
1
2
3
4
5. Consider the real plane R 2 . We agree that the two points (a t , a 2 ), (b lt b 2 )
suffice to describe a quadrilateral with corners at (0, 0), {a x , a 2 ), (b x , b 2 ), and
(a x + b x ,a 2 + b 2 ). (See Fig. 2.) Show that the area of this quadrilateral is
b \ b 2
(oi + bi, U2 + b 2 )
Fie. 2
3  Cofactors
93
Notice that the determinant can be positive or negative, and that it changes sign
if the first and second rows are interchanged. To interpret the value of the deter
minant as an area, we must either use the absolute value of the determinant or give
an interpretation to a negative area. We make the latter choice since to take the
absolute value is to discard information. Referring to Fig. 2, we see that the
direction of rotation from (a lf a 2 ) to {b lt b 2 ) across the enclosed area is the same as
the direction of rotation from the positive a^axis to the positive # 2 axis. To
interchange {a lt a 2 ) and (6 l9 b 2 ) would be to change the sense of rotation and the
sign of the determinant. Thus the sign of the determinant determines an orientation
of the quadrilateral on the coordinate system. Check the sign of the determinant
for choices of (a lf a 2 ) and (Jb ± , b 2 ) in various quadrants and various orientations.
6. (Continuation) Let E be an elementary transformation of R 2 onto itself.
E maps the vertices of the given quadrilateral onto the vertices of another quad
rilateral. Show that the area of the new quadrilateral is det E times the area of the
old quadrilateral.
7. Let x lt . . . , x n be a set of indeterminates. The determinant
1 x x CBj 2
«nl
X 2
1 x n x n
is called the Vandermonde determinant of order n.
(a) Show that Fis a polynomial of degree n — 1 in each indeterminate separately
and of degree n{n — l)/2 in all the indeterminates together.
(b) Show that, for each i <j, Kis divisible by x 6 — x t .
(c) Show that TJ (*, — x t ) is a polynomial of degree n — 1 in each in
l<i<3<n
determinate separately, and of degree n(n — l)/2 in all the indeterminates together.
(d) Show that V = JJ (x,  x<).
l<i<j<n
3 I Cofactors
For a given pair i, j, consider in the expansion for det A those terms which
have a H as a factor. Det A is of the form det A = a^A^ + (terms which do
not contain a u as a factor). The scalar A u is called the cofactor of a {j .
In particular, we see that A 1X = ^ (sgn n)a
2tt(2)
nn(n)
where this
sum includes all permutations n that leave 1 fixed. Each such n defines a
permutation 7/ on S' = {2, . . . , n} which coincides with tt on S. Since
no inversion of tt involves the element 1, we see that sgn tt = sgn tt'. Thus
A i} is a determinant, the determinant of the matrix obtained from A by
crossing out the first row and the first column of A.
94 Determinants, Eigenvalues, and Similarity Transformations  III
A similar procedure can be used to compute the cofactors A H . By a
sequence of elementary row and column operations of type III we can
obtain a matrix in which the element a {j is moved into row 1, column 1.
By applying the observation of the previous paragraph we see that the
cofactor A u is essentially the determinant of the matrix obtained by crossing
out the row and column containing the element a tj . Furthermore, we can
keep the other rows and columns in the same relative order if the sequence
of operations we use interchanges only adjacent rows or columns. It takes
i — 1 interchanges to move the element a ti into the first row, and it takes
j — 1 interchanges to move it into the first column. Thus A if is (— l) l 'i+>'i =
(_ iy+3 times the determinant of the matrix obtained by crossing out the
rth row and theyth column of A.
Each term in the expansion of det A contains exactly one factor from each
row and each column of A. Thus, for any given row of A each term of det A
contains exactly one factor from that row. Hence, for any given i,
det A = 2 a^A^. (3.1)
i
Similarly, for any given column of A each term of det A contains exactly one
factor from that column. Hence, for any given k,
det A = 2 a ik A ih . (3.2)
3
These expansions of a determinant according to the cofactors of a row
or column reduce the problem of computing an «th order determinant to
that of computing n determinants of order n — 1. We have already given
explicit expansions for determinants of orders 2 and 3, and the technique
of expansions according to cofactors enables us to compute determinants
of higher orders. The labor of evaluating a determinant of even quite
modest order is still quite formidable, however, and we make some suggestions
as to how the work can be minimized.
First, observe that if any row or column has several zeros in it, expansion
according to cofactors of that row or column will require the evaluation of
only those cofactors corresponding to nonzero elements. It is clear that
the presence of several zeros in any row or column would considerably
reduce the labor. If we are not fortunate enough to find such a row or
column, we can produce a row or column with a large number of zeros by
applying some elementary operations of type II. For example, consider
the determinant
det ,4 =
3 22 10
3 112
2234
115 2
3 I Cofactors
95
If the numbers appearing in the array were unwieldy, there would be no
choice but to wade in and make the best of it. The numbers in our example
are all integers, and we will not introduce fractions if we take advantage
of the l's that appear in the array. By Theorem 2.5, a sequence of elementary
operations of type II will not change the value of the determinant. Thus we
can obtain
1 17 4
det^ =
2 14 4
4 13 8
1
1
1 17
2
4
14 4
13 8
Now we face several options. We can expand the 3rd order determinants as
it stands; we can try the same technique again; or we can try to remove a
common factor from some row or column. We can remove the common
factor —1 from the second row and the common factor 4 from the third
column. Although 2 is factor of the second row, we cannot remove both
a 2 from the second row and a 4 from the third column. Thus we can obtain
det A = 4
1 17
14
13
1
17
1
= 4
3
31
6
47
3
31
= 4
6
47
—
1
If we multiply the elements in row i by the cofactors of the elements in
row k 7^ i, we get the same result as we would if the elements in row k were
equal to the elements in row /. Hence,
and
2 a it A M = for i ^ k,
2 a^A^ = for j ^ k.
(3.3)
(3.4)
The various relations we have developed between the elements of a matrix
and their cofactors can be summarized in the form
2 «.• A* = <5<* det A,
(3.5)
2 a a A iic = djh det A.
(3.6)
If A = [a {j ] is any square matrix and A tj is the cofactor of a ijy the matrix
[A tj ] T = adj A is called the adjunct of A. What we here call the "adjunct"
96
Determinants, Eigenvalues, and Similarity Transformations  III
is traditionallly called the "adjoint." Unfortunately, the term "adjoint"
is also used to denote a linear transformation that is not represented by the
adjoint (or adjunct) matrix. A new term is badly needed. We shall have a
use for the adjunct matrix only in this chapter. Thus, this unconventional
terminology will cause only a minor inconvenience and help to avoid con
fusion.
Theorem 3.1.
PROOF.
Aad]A = (adj A) • A = (det A) • /.
A • adj A = [a it ] • [A kl f =
2 a a A u
= (det A) • I.
(3.7)
(adj A)A = [A kl f
[««] =
I *
ik u ii
= (det A) I. a
(3.8)
Theorem 3.1 provides us with an effective technique for computing the
inverse of a nonsingular matrix. However, it is effective only in the sense
that the inverse can be computed by a prescribed sequence of steps. The
number of steps is large for matrices of large order, and it is not sufficiently
small for matrices of low order to make it a preferred technique. The method
described in Section 6 of Chapter II is the best method that is developed in
this text. In numerical analysis where matrices of large order are inverted,
highly specialized methods are available. But a discussion of such methods
is beyond the scope of this book.
A matrix A is nonsingular if and only if det A ^ 0, and in this case we
can see from the theorem that
s% =
— au s\.
det A
This is illustrated in
the following example.
" 1
2 3"
"3
5
1
A =
2
1 2
adj A =
2
5
4
_2
1 1_
*
4
5
3
"3 5
r
A~* = i
2 5
4
4 5 
3
(3.9)
The relations between the elements of a matrix and their cofactors lead
to a method for solving a system of n simultaneous equations in n unknowns
3 I Cofactors
97
when the equations are independent. Suppose we are given the system of
equations
J.a i§ x s =b it (i = l,2,...,n). (3.10)
3=1
The assumption that the equations are independent is expressed in the
condition that det A ^ 0, where A = [a it ]. Let A H be the cofactor of a^.
Then for a given k
n / n \ n i n \
2 A ik { 2,a tJ xA = 22 A ik a i Ax i
n
= 2 det 4 w a;,
Since det A y£ we see that
= det A z fc = 2 A ncK
3/i. —
.2, ^ifc^i
i=l
det A
(3.11)
(3.12)
The numerator can be interpreted as the cofactor expansion of the deter
minant of the matrix obtained by replacing the kth column of A by the
column of the b t . In this form the method is known as Cramer's rule.
Cramer's rule is convenient for systems of equations of low order, but
it fails if the system of equations is dependent or the number of equations
is different from the number of unknowns. Even in these cases Cramer's
rule can be modified to provide solutions. However, the methods we have
already developed are usually easier to apply, and the balance in their favor
increases as the order of the system of equations goes up and the nullity
increases.
EXERCISES
1. In the determinant
2 7 5 8
7125
10 4 2
3 61 2
find the cofactor of the "8" ; find the cofactor of the " 3."
2. The expansion of a determinant in terms of a row or column, as in formulas
(3.1) and (3.2), provides a convenient method for evaluating determinants. The
98
Determinants, Eigenvalues, and Similarity Transformations  III
amount of work involved can be reduced if a row or column is chosen in which
some of the elements are zeros. Expand the determinant
1
3
4
1
2
2
1
1
1
3
3
1
2
in terms of the cofactors of the third row.
3. It is even more convenient to combine an expansion in terms of cofactors
with the method of elementary row and column operations described in Section 2.
Subtract appropriate multiples of column 2 from the other columns to obtain
1
3
7
8
2
2
2
7
1
3
1
2
and expand this determinant in terms of cofactors of the third row.
4. Show that det (adj A) = (det A)" 1 .
5. Show that a matrix is nonsingular if and only if its adj A is also nonsingular.
6. Let A = [ciij] be an arbitrary n x n matrix and let adj A be the adjunct of A.
If X = (x lt . . . , x n ) and Y = (y lt . . . , y n ) show that
y y (adj A)X = 
For notation see pages 42 and 55.
V\
4 I The HamiltonCayley Theorem
Let p(x) = a m x m + • • • + a be a polynomial in an indeterminate x
with scalar coefficients a^ If A is an n x n matrix, by p(A) we mean the
matrix a m A m + a m _ x A m ~ x H + a I. Notice particularly that the
constant term a must be replaced by a I so that each term of p(A) will be
a matrix. No particular problem is encountered with matric polynomials of
this form since all powers of a single matrix commute with each other.
Any polynomial identity will remain valid if the indeterminate is replaced
4  The HamiltonCayley Theorem
99
by a matrix, provided any scalar terms are replaced by corresponding scalar
multiples of the identity matrix.
We may also consider polynomials with matric coefficients. To make
sense, all coefficients must be matrices of the same order. We consider
only the possibility of substituting scalars for the indeterminate, and in all
manipulations with such polynomials the matric coefficients commute with
the powers of the indeterminate. Polynomials with matric coefficients can
be added and multiplied in the usual way, but the order of the factors
is important in multiplication since the coefficients may not commute. The
algebra of polynomials of this type is not simple, but we need no more than
the observation that two polynomials with matric coefficients are equal if
and only if they have exactly the same coefficients.
We avoid discussing the complications that can occur for polynomials
with matric coefficients in a matric variable.
Now we should like to consider matrices for which the elements are
polynomials. If F is the field of scalars for the set of polynomials in the
indeterminate x, let K be the set of all rational functions in x; that is, the
set of all permissible quotients of polynomials in x. It is not difficult to show
that K is a field. Thus a matrix with polynomial components is a special
case of a matrix with elements in K.
From this point of view a polynomial with matric coefficients can be
expressed as a single matrix with polynomial components. For example,
x 2 + 2 2x  1
x 2  2x + 1 2x 2 + 1.
Conversely, a matrix in which the elements are polynomials in an indeter
minate x can be expanded into a polynomial with matric coefficients. Since
polynomials with matric coefficients and matrices with polynomial compo
nents can be converted into one another, we refer to both types of expressions
as polynomial matrices.
Definition. If A is any square matrix, the polynomial matrix A — xl —
C is called the characteristic matrix of A .
C has the form
r i °i
r o 2i
[2
11
1 2,
x 2 +
2 0_
x +
1
1_
—
flu — X
floo X
[_ "nl
'n2
«2„
(4.1)
a„„ — x
100 Determinants, Eigenvalues, and Similarity Transformations  III
The determinant of C is a polynomial det C = f(x) = k n x n + k n _ x x n  x +
 • • + k of degree n; it is called the characteristic polynomial of A. The
equation /(x) = is called the characteristic equation of A. First, we should
observe that the coefficient of x n in the characteristic polynomial is (— l) n ,
the coefficient of x n ~ x is (— l) n_1 £ 4 n =1 a u> and the constant term k = det A.
Theorem 4.1. {HamiltonCayley theorem). If A is a square matrix and
f{x) is its characteristic polynomial, then j (A) = 0.
proof. Since C is of order n, adj C will contain polynomials in x of degree
not higher than n — 1. Hence adj C can be expanded into a polynomial
with matric coefficients of degree at most n — 1 :
adj C = C n _ x x n ~ x + C„_ 2 x" 2 + • • • + C x x + C (4.2)
where each C t is a matrix with scalar elements.
By Theorem 3.1 we have
adjC C = det C /=/(>;)/
= adj C • {A  xl) = (adj C)A  (adj C)x. (4.3)
Hence,
k n Ix n + k n _Jx n  x \ + kjx + k I
+ CV^a;" 1 + • • • + C^x + C Q A. (4.4)
The expressions on the two sides of this equality are n X n polynomial
matrices. Since two polynomial matrices are equal if and only if the cor
responding coefficients are equal, (4.4) is equivalent to the following set
of matric equations:
kn* = C n _ x
k n \l = C n 2 + C n _ x A
(4.5)
k I = C ^4.
Multiply each of these equations by A n , A n ~ x , . . . , A, I from the right,
respectively, and add them. The terms on the right side will cancel out
leaving the zero matrix. The terms on the left add up to
k n A n + k n _ x A n ~ x + .. + h x A + k Q I=f(A) = 0. □ (4.6)
The equation m(x) = of lowest degree which A satisfies is called the
minimum equation (or minimal equation) for A ; m(x) is called the minimum
polynomial for A. Since A satisfies its characteristic equation the degree
of m(x) is not more than n. Since a linear transformation and any matrix
4  The HamiltonCayley Theorem 101
representing it satisfy the same relations, similar matrices satisfy the same
set of polynomial equations. In particular, similar matrices have the same
minimum polynomials.
Theorem 4.2. If g(x) is any polynomial with coefficients in F such that
g(A) — 0, then g(x) is divisible by the minimum polynomial for A. The
minimum polynomial is unique except for a possible nonzero scalar factor.
proof. Upon dividing g(x) by m(x) we can write g{x) in the form
g(x) = m{x) • q{x) + r(x), (4.7)
where q(x) is the quotient polynomial and r{x) is the remainder, which is
either identically zero or is a polynomial of degree less than the degree of
m(x). Ifg(x) is a polynomial such that g(A) = 0, then
g{A) = = m(A) • q(A) + r(A) = r{A). (4.8)
This would contradict the selection of m(x) as the minimum polynomial
for A unless the remainder r(x) is identically zero. Since two polynomials
of the same lowest degree must divide each other, they must differ by a
scalar factor. □
As we have pointed out, the elements of adj C are polynomials of degree at
most n — 1. Let g(x) be the greatest common divisor of the elements of
adj C. Since adj C ■ C =/(*)/, g(x) divides /(x).
f( z )
Theorem 4.3. h(x) = —— is the minimum polynomial for A.
g(x)
proof. Let adj C = g(x)B where the elements of B have no nonscalar
common factor. Since adj C • C = f{x)Iwe have h(x) • g(x)I = g(x)BC. Since
g(x) j£ this yields
BC = h(x)I. (4.9)
Using B in place of adj C we can repeat the argument used in the proof of the
HamiltonCayley theorem to deduce that h{A) = 0. Thus h(x) is divisible
by m(x).
On the other hand, consider the polynomial m(x) — m(y). Since it is a
sum of terms of the form c^ — y*), each of which is divisible by y — x,
m(x) — m(y) is divisible by y — x:
m(x)  m(y) = (y  x) • k(x, y). (4.10)
Replacing x by xl and y by A we have
m(xl)  m(A) = m(x)I = (A  xl) • k(xl, A) = C ■ k(xl, A). (4.1 1)
Multiplying by adj C we have
m(x) adj C = (adj C)C • k(xl, A) = f(x) • k(xl, A). (4.12)
102 Determinants, Eigenvalues, and Similarity Transformations  III
Hence,
m(x) ■ g{x)B = h{x) ■ g(x) • k(xl, A), (4. 13)
m(x)B = h(x) • k(xl, A). (4.14)
or
Since h{x) divides every element of m{x)B and the elements of B have no
nonscalar common factor, h(x) divides m(x). Thus, h{x) and m{x) differ
at most by a scalar factor. □
Theorem 4.4. Each irreducible factor of the characteristic polynomial f(x)
of A is also an irreducible factor of the minimum polynomial m(x).
proof. As we have seen in the proof of the previous theorem
m(x)I = C • k(xl, A).
Thus
det m(x)I = [m(x)] n = det C • det k(xl, A)
= f(x) det k(xl, A). (4.15)
We see then that every irreducible factor off(x) divides [m(x)] n , and therefore
m(x) itself. □
Theorem 4.4 shows that a characteristic polynomial without repeated
factors is also the minimum polynomial. As we shall see, it is the case in
which the characteristic polynomial has repeated factors that generally causes
trouble.
We now ask the converse question. Given the polynomial f(x) =
{—\) n x n + k n _^ x x n ~ x + • • • + k , does there exist an n x n matrix A for
which f(x) is the minimum polynomial ?
Let A = {x lt . . . , <x n } be any basis and define the linear transformation
a by the rules
oiui) = a i+1 for i < n, (4.16)
and
(l) n cr(a n ) = k oc x  k x cc 2 A: n _ x a n .
It follows directly from the definition of a that
/(<0(«i) = (l)MaJ + & n _ x a n + • • • + k t 0L 2 + k a x = 0. (4.17)
For any other basis element we have
/(*)(«*) =A«)[<y J  1 M] = o^[f{a)M] = 0. (4.18)
Since f(a) vanishes on the basis elements f(a) = and any matrix repre
senting a satisfies the equation f(x) = 0.
4  The HamiltonCayley Theorem 103
On the other hand, a cannot satisfy an equation of lower degree because
the corresponding polynomial in a applied to a x could be interpreted as
a relation among the basis elements. Thus, f(x) is a minimum polynomial
for a and for any matrix representing a. Since f(x) is of degree n, it must
also be the characteristic polynomial of any matrix representing a.
With respect to the basis A the matrix representing a is
A =
1
1
(1)%
 (l) n *i
(l) n k 2
(4.19)
1 (1)^^
A is called the companion matrix off(x).
Theorem 4.5. f(x) is a minimum polynomial for its companion matrix. □
EXERCISES
1. Show that x 3 + 39z  90 is the characteristic polynomial for the matrix
"0 90"
1 39
1
2. Find the characteristic polynomial for the matrix
~2 2 3"
1 1 1
1 3 1
and show by direct substitution that this matrix satisfies its characteristic equation.
3. Find the minimum polynomial for the matrix
^3 2 2"
1 4 1
2 4 1
4. Write down a matrix which has x 4 + 3* 3 + 2x 2  x + 6 = as its minimum
equation.
104 Determinants, Eigenvalues, and Similarity Transformations  III
5. Show that if the matrix A satisfies the equation x 2 + x + 1 =0, then A is
nonsingular and the inverse A 1 is expressible as a linear combination of A and /.
6. Show that no real 3x3 matrix satisfies x 2 + 1 =0. Show that there are
complex 3x3 matrices which do. Show that there are real 2x2 matrices that
satisfy the equation.
7. Find a 2 x 2 matrix with integral elements satisfying the equation x 3 — 1 =0,
but not satisfying the equation x — 1 =0.
8. Show that the characteristic polynomial of
7 44"
4 8 1
4 1 8
is not its minimum polynomial. What is the minimum polynomial ?
5 I Eigenvalues and Eigenvectors
Let a be a linear transformation of V into itself. It is often useful to find
subspaces of V in which a also acts as a linear transformation. If W is such
a subspace, this means that cr(VV) <= W. A subspace with this property is
called an invariant subspace of V under a. Generally, the problem of deter
mining the properties of a on V can be reduced to the problem of determining
the properties of a on the invariant subspaces.
The simplest and most restricted case occurs when an invariant subspace
W is of dimension 1. In that case, let {a x } be a basis for W. Then, since
o{v.i) e W, there is a scalar X x such that o{vl x ) = X x <x x . Also for any a e W,
a = a x a x and hence c(a) = a x a(a x ) = a x X x y. x = X x a. In some sense the
scalar X x is characteristic of the invariant subspace W; a stretches every
vector in W by the factor X x .
In general, a problem of finding those scalars X and associated vectors
I for which o(£) = A is called an eigenvalue problem. A nonzero vector
£ is called an eigenvector of a if there exists a scalar X such that c(£) = X£.
A scalar X is called an eigenvalue of a if there exists a nonzero vector £
such that <r(£) = Xg. Notice that the equation <r() = X£ is an equation in
two variables, one of which is a vector and the other a scalar. The solution
1 = and X any scalar is a solution we choose to ignore since it will not
lead to an invariant subspace of positive dimension. Without further
conditions we have no assurance that the eigenvalue problem has any other
solutions.
A typical and very important eigenvalue problem occurs in the solution
of partial differential equations of the form
d 2 u d 2 u
dx* dy 2 ~ '
5  Eigenvalues and Eigenvectors 105
subject to the boundary conditions that w(0, y) = u(tt, y) = 0,
lim u(x, y) = 0, and u(x, 0) = f(x)
y ►oo
where f(x) is a given function. The standard technique of separation of
variables leads us to try to construct a solution which is a sum of functions
of the form XY where X is a function of x alone and Y is a function of
y alone. For this type of function, the partial differential equation becomes
dx 2 dy 2
Since
1 .^!l = _ A .<!*x
Y dy 2 X dx 2
is a function of x alone and also a function of y alone, it must be a constant
(scalar) which we shall call k 2 . Thus we are trying to solve the equations
—  —k 2 X —  k 2 Y
dx 2 ~ kX > dy 2~ k Y 
These are eigenvalue problems as we have defined the term. The vector
space is the space of infinitely differentiable functions over the real numbers
and the linear transformation is the differential operator d 2 /dx 2 .
For a given value of k 2 (k > 0) the solutions would be
X = a x cos kx + a 2 sin kx,
Y = a 3 e~ kv + a 4 e*».
The boundary conditions w(0, y) = and liny^ u(x, y) = imply that
a i = a \ = 0. The most interesting condition for the purpose of this example
is that the boundary condition u(tt, y) = implies that k is an integer.
Thus, the eigenvalues of this eigenvalue problem are the integers, and the
corresponding eigenfunctions (eigenvectors) are of the form a k e kv sin kx.
The fourth boundary condition leads to a problem in Fourier series; the
problem of determining the a k so that the series
oo
2 a k sin kx
represents the given function f(x) for < x < n.
Although the vector space in this example is of infinite dimension, we
restrict our attention to the eigenvalue problem in finite dimensional vector
spaces. In a finite dimensional vector space there exists a simple necessary
and sufficient condition which determines the eigenvalues of an eigenvalue
problem.
The eigenvalue equation can be written in the form (a — A)(£) = 0.
We know that there exists a nonzero vector  satisfying this condition if
106 Determinants, Eigenvalues, and Similarity Transformations  III
and only if a — X is singular. Let A = {a l5 . . . , a n } be any basis of V and
let A = [a i} ] be the matrix representing a with respect to this basis. Then
A — XI = C(X) is the matrix representing a — X. Since A — XI is singular
if and only if det {A — XI) = f(X) = 0, we see that we have proved
Theorem 5.1. A scalar X is an eigenvalue of a if and only if it is a solution
of the characteristic equation of a matrix representing a. □
Notice that Theorem 5.1 applies only to scalars. In particular a solution
of the characteristic equation which is not a scalar is not an eigenvalue. For
example, if the field of scalars is the field of real numbers, then nonreal
complex solutions of the characteristic equation are not eigenvalues. In the
published literature on matrices the terms "proper values" and "characteristic
values" are also used to denote what we have called eigenvalues. But,
unfortunately, the same terms are often also applied to the solutions of the
characteristic equation. We call the solutions of the characteristic equation
characteristic values. Thus, a characteristic value is an eigenvalue if and only
if it is also in the given field of scalars. This distinction between eigenvalues
and characteristic values is not standard in the literature on matrices, but we
hope this or some other means of distinguishing between these concepts will
become conventional.
In abstract algebra a field is said to be algebraically closed if every poly
nomial with coefficients in the field factors into linear factors in the field.
The field of complex numbers is algebraically closed. Though many proofs
of this assertion are known, none is elementary. It is easy to show that
algebraically closed fields exist, but it is not easy to show that a specific field is
algebraically closed.
Since for most applications of concepts using eigenvalues or characteristic
values the underlying field is either rational, real or complex, we content
ourselves with the observation that the concepts, eigenvalue and characteristic,
value, coincide if the underlying field is complex, and do not coincide if the
underlying field is rational or real.
The procedure for finding the eigenvalues and eigenvectors of a is fairly
direct. For some basis A = {<x x , . . . , <x n }, let A be the matrix representing a.
Determine the characteristic matrix C(x) = A — xl and the characteristic
equation det {A — XI) = f(x) = 0. Solve the characteristic equation. (It is
this step that presents the difficulties. The characteristic equation may have
no solution in F. In that event the eigenvalue problem has no solution.
Even in those cases where solutions exists, finding them can present practical
difficulties.) For each solution X off(x) = 0, solve the system of homogene
ous equations
(A  XI)X = C(X) X=0. (5.1)
5  Eigenvalues and Eigenvectors 107
Since this system of equations has positive nullity, nonzero solutions exist
and we should use the Hermite normal form to find them. All solutions
are the representations of eigenvectors corresponding to a.
Generally, we are given the matrix A rather than a itself, and in this case
we regard the problem as solved when the eigenvalues and the representations
of the eigenvectors are obtained. We refer to the eigenvalues and eigenvectors
of a as eigenvalues and eigenvectors, respectively, of A.
Theorem 5.2. Similar matrices have the same eigenvalues and eigenvectors.
proof. This follows directly from the definitions since the eigenvalues
and eigenvectors are associated with the underlying linear transformation. □
Theorem 5.3. Similar matrices have the same characteristic polynomial.
proof. Let A and A' = P~ X AP be similar. Then
det (A'  xl) = det (P^AP  xl) = det {P~ X {A  xI)P) = detP 1
det {A  xl) det P = det (A  xl) =f(x). □
We call the characteristic polynomial of any matrix representing a the
characteristic polynomial of o. Theorem 5.3 shows that the characteristic
polynomial of a linear transformation is uniquely defined.
Let S(X) be the set of all eigenvectors of a corresponding to X, together
with 0.
Theorem 5.4. S(A) is a subspace of V.
proof. If a and (3 e S(X), then
a(aa. + bp) = ao{v.) + ba{p)
= aXcn + bXfi
= X(a<x + bp). (5.2)
Hence, ace + bp e S(A) and S(X) is a subspace. □
We call S(X) the eigenspace of a corresponding to X, and any subspace
of S(X) is called an eigenspace of a.
The dimension of S(X) is equal to the nullity of C(X), the characteristic
matrix of A with X substituted for the indeterminate x. The dimension of
S(X) is called the geometric multiplicity of X. We have shown that X is also
a solution of the characteristic equation f(x) = 0. Hence, (x — A) is a
factor of f(x). If (x — X) k is a factor of f(x) while (x — X) k+l is not, X is a
root of f(x) = of multiplicity k. We refer to this multiplicity as the
algebraic multiplicity of X.
Theorem 5.5. The geometric multiplicity of X does not exceed the algebraic
multiplicity of X.
proof. Since the geometric multiplicity of X is defined independently of
any matrix representing a and the characteristic equation is the same for all
108 Determinants, Eigenvalues, and Similarity Transformations  III
matrices representing a it will be sufficient to prove the theorem for any
particular matrix representing a. We shall choose the matrix representing
a so that the assertion of the theorem is evident. Let r be the dimension
of S(K) and let {f 1} . . . , £ r } be a basis of S{X). This linearly independent set
can be extended to a basis {£ 1? . . . , £J of V. Since o(£ t ) = A!< for / < r,
the matrix A representing a with respect to this basis has the form
A =
'X
I
«l,r+l
a
*2,r+l
1 a
r,r+l
a
r+l,r+l
a.
(5.3)
From the form of A it is evident that det {A — xl)=f(x) is divisible by
(x _ X) r . Therefore, the algebraic multiplicity of I is at least r, which is the
geometric multiplicity. □
Theorem 5.6. If the eigenvalues X x , . . . , X s are all different and {f ls ...,£,}
is a set of eigenvectors, ^ corresponding to X t , then the set {£ l5 . . . , £,} is
linearly independent.
proof. Suppose the set is dependent and that we have reordered the
eigenvectors so that the first k eigenvectors are linearly independent and
the last s — k are dependent on them. Then
k
is = 2 a &i
where the representation is unique. Not all a t = since  s ^ 0. Upon
applying the linear transformation a we have
There are two possibilities to be considered. If X s = 0, then none of the
A; for / < k is zero since the eigenvalues are distinct. This would imply
5  Eigenvalues and Eigenvectors 109
that {i u . . . , i k j is linearly dependent, contrary to assumption. If X s ^ 0,
then
k 1
<i A s
Since not all a t = and XJX S ^ 1, this would contradict the uniqueness of
the representation of £ s . Since we get a contradiction in any event, the
set {£ x , ... , f J must be linearly independent. □
EXERCISES
1. Show that X = is an eigenvalue of a matrix A if and only if A is singular.
2. Show that if £ is an eigenvector of cr, then I is also an eigenvector of a n for
each « > 0. If X is the eigenvalue of a corresponding to f , what is the eigenvalue
of a n corresponding to £ ?
3. Show that if £ is an eigenvector of both a and r, then £ is also an eigenvector
of aa{ for aeF) and ct + t. If A x is the eigenvalue of a and A 2 is the eigenvalue of
t corresponding to I, what are the eigenvalues of aa and a + T ?
4. Show, by producing an example, that if X x and A 2 are eigenvalues of <r x and a 2 ,
respectively, it is not necessarily true that X 1 + A 2 is an eigenvalue of a t + <r 2 .
5. Show that if £ is an eigenvector of a, then it is also an eigenvector of p(o)
where p{x) is a polynomial with coefficients in F. If X is an eigenvalue of a corre
sponding to I, what is the eigenvalue of p{a) corresponding to f ?
6. Show that if a is nonsingular and X is an eigenvalue of a, then A" 1 is an
eigenvalue of <r _1 . What is the corresponding eigenvector?
7. Show that if every vector in V is an eigenvector of a, then a is a scalar trans
formation.
8. Let P n be the vector space of polynomials of degree at most n — 1 , and let D
be the differentiation operator; that is D(t k ) = kt* 1 . Determine the characteristic
polynomial for D. From your knowledge of the differentiation operator and net
using Theorem 4.3, determine the minimum polynomial for D. What kind of
differential equation would an eigenvector of D have to satisfy? What are the
eigenvectors of D ?
9. Let A = [a^]. Show that if ^ «« = c independent of /, then I = (1 , 1 , . . . ,
1) is an eigenvector. What is the corresponding eigenvalue?
10. Let W be an invariant subspace of V under a, and let A = {a l5 . . . , a n } be a
basis of V such that {a x , . . . , a] J is a basis of W. Let ^ = [a i3 ] be the matrix
representing a with respect to the basis A. Show that all elements in the first k
columns below the fcth row are zeros.
11. Show that if X 1 and A 2 ^ X x are eigenvalues of a x and £ x and £ 2 are eigen
vectors corresponding to X x and X 2 , respectively, then f x + £ 2 is not an eigenvector.
12. Assume that {$ lf . . . ,  lr } are eigenvectors with distinct eigenvalues. Show
that 2i =1 a t £j is never an eigenvector unless precisely one coefficient is nonzero.
110
Determinants, Eigenvalues, and Similarity Transformations  HI
13. Let A be an n x n matrix with eigenvalues A lt A 2 , . . . , A„. Show that if A
is the diagonal matrix
"li • • • ~~
A =
A 2
I
and P = [pij] is the matrix in which column j is the MtupIe representing an eigen
vector corresponding to A 3 , then AP = PA.
14. Use the notation of Exercise 13. Show that if A has n linearly independent
eigenvalues, then eigenvectors can be chosen so that P is nonsingular. In this case
p^AP = A.
6 I Some Numerical Examples
Since we are interested here mainly in the numerical procedures, we
start with the matrices representing the linear transformations and obtain
their eigenvalues and the representations of the eigenvectors.
Example 1. Let
"I 2 2"
A =
2 2 2
3 6 6
The first step is to obtain the characteristic matrix
_1 _ x 2 2
C{x) = 2 2 x 2
_3 _6 6 
and then the characteristic polynomial
detC(a:)= (* + 2)(a + 3)a\
Thus the eigenvalues of A are A x = —2, A 2 = —3, and A 3 = 0. The next
steps are to substitute, successively, the eigenvalues for x in the characteristic
matrix. Thus we have
1 2 2~
C(2) =
2 4 2
3 6 4
6  Some Numerical Examples jjj
The Hermite normal form obtained from C(— 2) is
1 2
1
_0 0_
The components of the eigenvector corresponding to X x = — 2 are found
by solving the equations
x x + 2x 2 =0
x 3 = 0.
Thus (2, —1,0) is the representation of an eigenvector corresponding to
X x \ for simplicity we shall write ^ = (2, 1,0), identifying the vector
with its representation.
In a similar fashion we obtain
"222
C(3)= 2 5 2
__3 _6 3
From C(— 3) we obtain the Hermite normal form
"1 1"
1
0.
and hence the eigenvector £ 2 = (1,0, —1).
'1 2 2"
C(0)= 2 2 2
Similarly, from
_3 _6 6
we obtain the eigenvector £ 3 = (0, 1, —1).
By Theorem 5.6 the three eigenvectors obtained for the three different
eigenvalues are linearly independent.
Example 2. Let
A =
From the characteristic matrix
C(x) =
" 1
i r
1
3 1
_l
2 0_
1  X
i r
1
3 — x 1
1
2
—x
112 Determinants, Eigenvalues, and Similarity Transformations  III
we obtain the characteristic polynomial det C(x) = — (x — l) 2 (x — 2).
Thus we have just two distinct eigenvalues; X x = K = 1 with algebraic
multiplicity two, and /l 3 = 2.
Substituting A x for x in the characteristic matrix we obtain
C(l) =
"
1
1
1 1"
2 1
2 1
The corresponding Hermite normal form is
1 1"
1 1
Thus it is seen that the nullity of C(l) is 1 . The eigenspace S(l) is of dimension
1 and the geometric multiplicity of the eigenvalue 1 is 1. This shows that the
geometric multiplicity can be lower than the algebraic multiplicity. We obtain
& = (1,1,1).
The eigenvector corresponding to X z = 2 is £ 3 = (0, 1, 1).
EXERCISES
For each of the following matrices find all the eigenvalues and as many linearly
independent eigenvectors as possible.
1.
"2 4"
5 3
3.
"1 2"
2 2
5.
7.
'4
9 0"

2 8
7
" 7
4 4"
4
8 1
4
1 
8
2.
6.
3
2
2"
3
1
V2
V2
4
3
2
2"
1
4
1
2
4 
1
2
—i
0"
/
2
9
3
7  Similarity
113
7 I Similarity
Generally, for a given linear transformation a we seek a basis for which
the matrix representing a has as simple a form as possible. The simplest
form is that in which the elements not on the main diagonal are zero, a
diagonal matrix. Not all linear transformations can be represented by
diagonal matrices, but relatively large classes of transformations can be
represented by diagonal matrices, and we seek conditions under which
such a representation exists.
Theorem 7.1. A linear transformation a can be represented by a diagonal
matrix if and only if there exists a basis consisting of eigenvectors of a.
proof. Suppose there is a linearly independent set X = {f l9 . . . , f } of
eigenvectors and that {A l5 . . . , X n ) are the corresponding eigenvalues.
Then a(^) = A^ so that the matrix representing a with respect to the
basis X has the form
X x
L
(7.1)
that is, cr is represented by a diagonal matrix.
Conversely, if a is represented by a diagonal matrix, the vectors in that
basis are eigenvectors. □
Usually, we are not given the linear transformation a directly. We are
given a matrix A representing a with respect to an unspecified basis. In
this case Theorem 7.1 is usually worded in the form: A matrix^ is similar
to a diagonal matrix if and only if there exist n linearly independent eigen
vectors of A. In this form a computation is required. We must find the matrix
P such that P~ X AP is a diagonal matrix.
Let the matrix A be given; that is, A represents a with respect to some
basis A = {a x , . . ., aj. Let f y = 2?=i/>««* be the representations of the
eigenvectors of A with respect to A. Then the matrix A' representing a
with respect to the basis X = {g lt ... , ^ n } is P^AP = A'. By Theorem
7.1, A' is a diagonal matrix.
In Example 1 of Section 6, the matrix
A =
1 2
2 2
3 6
114
Determinants, Eigenvalues, and Similarity Transformations  III
2
1
0"
" 1
1
1"
1
1
pl =
1
2
2
1
1
»
1
2
1
1
1
1
1
3
1
1
2
has three linearly independent eigenvectors, £ x = (2, — 1,0), £ 2 = 0> °> — 0»
and f s = (0, 1, —1). The matrix of transition P has the components of
these vectors written in its columns :
P =
The reader should check that P~ X AP is a diagonal matrix with the eigenvalues
appearing in the main diagonal.
In Example 2 of Section 6, the matrix
A =
has one linearly independent eigenvector corresponding to each of its two
eigenvalues. As there are no other eigenvalues, there does not exist a set of
three linearly independent eigenvectors. Thus, the linear transformation
represented by this matrix cannot be represented by a diagonal matrix;
A is not similar to a diagonal matrix.
Corollary 7.2. If a can be represented by a diagonal matrix D, the elements
in the main diagonal of D are the eigenvalues of a. U
Theorem 7.3. If an n X n matrix has n distinct eigenvalues, then A is
similar to a diagonal matrix.
proof. By Theorem 5.6 the n eigenvectors corresponding to the n eigen
values of A are linearly independent and form a basis. By Theorem 7.1
the matrix representing the underlying linear transformation with respect
to this basis is a diagonal matrix. Hence, A is similar to a diagonal matrix. □
Theorem 7.3 is quite practical because we expect the eigenvalues of a
randomly given matrix to be distinct; however, there are circumstances
under which the theorem does not apply. There may not be n distinct
eigenvalues, either because some have algebraic multiplicity greater than
one or because the characteristic equation does not have enough solutions in
the field. The most general statement that can be made without applying
more conditions to yield more results is
Theorem 7.4. A necessary and sufficient condition that a matrix A be
similar to a diagonal matrix is that its minimum polynomial factor into distinct
linear factors with coefficients in F.
7  Similarity 115
proof. Suppose first that the matrix A is similar to a diagonal matrix D.
By Theorem 5.3, A and D have the same characteristic polynomial. Since
D is a diagonal matrix the elements in the main diagonal are the solutions
of the chracteristic equation and the characteristic polynomial must factor
into linear factors. By Theorem 4.4 the minimum polynomial for A must
contain each of the linear factors of f(x), although possibly with lower
multiplicity. It can be seen, however, either from Theorem 4.3 or by direct
substitution, that D satisfies an equation without repeated factors. Thus,
the minimum polynomial for A has distinct linear factors.
On the other hand, suppose that the minimum polynomial for A is
m(x) = {x  X x ){x  A 2 ) • • • {x  X p ) with distinct linear factors. Let
M* be the kernel of a — X t . The nonzero vectors in M t are the eigenvectors
of a corresponding to X t . It follows from Theorem 5.6 that a nonzero
vector in M t cannot be expressed as a sum of vectors in ^ . M,. Hence,
the sum M x + M 2 + ( M p is direct.
Let v t = dim M„ that is, v i is the nullity of a  X t . Since M x • • • ©
M p c: V we have v x + • • • + v v ^ n. By Theorem 1.5 of Chapter II
dim (a — XJV = n — v t = Pi , By another application of the same theorem
we have dim (a  A,){(cr _ A 2 )V} > Pi  Vj = n  (v, + Vj ).
Finally, by repeated application of the same ideas we obtain =
dim m(a)V > n  (v x + • • • + Vp ). Thus, v x + • • • + v p = n. This shows
that M 1 ®"@M 9 = V. Since every vector in Visa linear combination of
eigenvectors, there exists a basis of eigenvectors. Thus, A is similar to a
diagonal matrix. □
Theorem 7.4 is important in the theory of matrices, but it does not provide
the most effective means for deciding whether a particular matrix is similar
to diagonal matrix. If we can solve the characteristic equation, it is easier
to try to find the n linearly independent eigenvectors than it is to use Theorem
7.4 to ascertain that they do or do not exist. If we do use this theorem and
are able to conclude that a basis of eigenvectors does exist, the work done in
making this conclusion is of no help in the attempt to find the eigenvectors.
The straightforward attempt to find the eigenvectors is always conclusive
On the other hand, if it is not necessary to find the eigenvectors, Theorem 7.4
can help us make the necessary conclusion without solving the characteristic
equation.
For any square matrix A = [a tj ], Tr(A) = 2? =1 a u is called the trace of A.
It is the sum of the elements in the diagonal of A. Since Tr{AB) =
ItidU °iM = IjLiCZTi V«) = Tr(BA),
Tr(/>MP) = Tr(APPi) = Tr(A). (7.2)
This shows that the trace is invariant under similarity transformations;
116
Determinants, Eigenvalues, and Similarity Transformations  III
that is, similar matrices have the same trace. For a given linear transforma
tion a of V into itself, all matrices representing a have the same trace. Thus
we can define Tr(cr) to be the trace of any matrix representing a.
Consider the coefficient of x n ~ x in the expansion of the determinant of the
characteristic matrix,
#22 "^
#m
a 2n
— X
(7.3)
The only way an x n ~ x can be obtained is from a product of n — 1 of the
diagonal elements, multiplied by the scalar from the remaining diagonal
element. Thus, the coefficient of x n ~ x is ( l)"" 1 2?=i a u , or ( l) B  1 Tr(,4).
If/(at) = det (A — xl) is the characteristic polynomial of A, then det A =
/(0) is the constant term of /(a;). If /(a) is factored into linear factors in the,
form
f{x) = ( l) n (x  Itf^x  A 2 )'» ■•■{x X P Y\ (7.4)
the constant term is YLLi K Thus det A is the P roduct of the characteristic
values (each counted with the multiplicity with which it is a factor of the
characteristic polynomials). In a similar way it can be seen that Tr(^) is the
sum of the characteristic values (each counted with multiplicity).
We have now shown the existence of several objects associated with a
matrix, or its underlying linear transformation, which are independent of
the coordinate system. For example, the characteristic polynomial, the
determinant, and the trace are independent of the coordinate system.
Actually, this list is redundant since det A is the constant term of the char
acteristic polynomial, and Tr(^) is ( 1)" _1 times the coefficient of a;" 1 of the
characteristic polynomial. Functions of this type are of interest because they
contain information about the linear transformation, or the matrix, and they
are sometimes rather easy to evaluate. But this raises a host of questions.
What information do these invariants contain? Can we find a complete
list of nonredundant invariants, in the sense that any other invariant can
be computed from those in the list? While some partial answers to these
questions will be given, a systematic discussion of these questions is beyond
the scope of this book.
7  Similarity 117
Theorem 7.5. Let V be a vector space with a basis consisting of eigen
vectors of g. IfW is any subspace of V invariant under a, then W also has a
basis consisting of eigenvectors of a.
proof. Let a be any vector in W. Since V has a basis of eigenvectors
of a, a can be expressed as a linear combination of eigenvectors of a. By
disregarding terms with zero coefficients, combining terms corresponding
to the same eigenvalue, and renaming a term like a^, where £ t is an eigen
vector and a t ^ 0, as an eigenvector with coefficient 1 , we can represent a in
the form
r
a = 2 £<»
where the £ f are eigenvectors of a with distinct eigenvalues. Let X { be the
eigenvalue corresponding to  t . We will show that each ^ £ W.
(a — l 2 )(oc — ^3) ' • ' (c — X r )(<x) is in W since W is invariant under a,
and hence invariant under a — X for any scalar X. But then (a — X 2 )(a — X 3 )
• • • (a — A r )(a) = (A x — ^X^ — X 3 ) • • • (X x — A r ) x e W, and  x £ W since
(A a — X 2 )(X 1 — X 3 ) • • • {X x — X r ) ?± 0. A similar argument shows that each
f,eW.
Since this argument applies to any a £ W, W is spanned by eigenvectors
of (T. Thus, W has a basis of eigenvectors of a. D
Theorem 7.6. Let V be a vector space over C, the field of complex numbers.
Let a be a linear transformation of V into itself V has a basis of eigenvectors
for a if and only if for every subspace S invariant under a there is a subspace T
invariant under a such that V = S © 7.
proof. The theorem is obviously true if V is of dimension 1. Assume
the assertions of the theorem are correct for spaces of dimension less than n,
where n is the dimension of V.
Assume first that for every subspace S invariant under a there is a com
plementary subspace T also invariant under a. Since V is a vector space over
the complex numbers a has at least one eigenvalue X ± . Let a 2 be an eigenvector
corresponding to Xj. The subspace S 2 = <a x > is then invariant under a.
By assumption there is a subspace 7^ invariant under a such that V = S x © Tj.
Every subspace S 2 of T x invariant under Ra is also invariant under a. Thus
there exists a subspace T 2 of V invariant under a such that V = S 2 © T 2 .
Now S 2 c T x and 7 a = S 2 © (T 2 n TJ. (See Exercise 15, Section 14.) Since
T 2 n 7 X is invariant under c, and therefore under Ra, the induction
assumption holds for the subspace T x . Thus, Tj has a basis of eigenvectors,
and by adjoining a x to this basis we obtain a basis of eigenvectors of V.
Now assume there is a basis of V consisting of eigenvectors of a. By
theorem 7.5 any invariant subspace S has a basis of eigenvectors. The method
118 Determinants, Eigenvalues, and Similarity Transformations  III
of proof of Theorem 2.7 of Chapter I (the Steinitz replacement theorem)
will yield a basis of V consisting of eigenvectors of o, and this basis will con
tain the basis of S consisting of eigenvectors. The eigenvectors adjoined
will span a subspace 7", and this subspace will be invariant under a and
complementary to S. □
EXERCISES
1. For each matrix A given in the exercises of Section 6 find, when possible,
a nonsingular matrix P for which P~ 1 AP is diagonal.
"1 c
where c ?* is not similar to a diagonal matrix
2. Show that the matrix
1
3. Show that any 2x2 matrix satisfying x 2 + 1 = is similar to the matrix
"0 1"
1
4. Show that if A is nonsingular, then AB is similar to BA.
5. Show that any two projections of the same rank are similar.
*8 I The Jordan Normal Form
A normal form that is obtainable in general when the field of scalars is
the field of complex numbers is known as the Jordan normal form. An
application of the Jordan normal form to power series of matrices and sys
tems of linear differential equations is given in the chapter on applications.
Except for these applications this section can be skipped without penalty.
We assume that the field of scalars is the field of complex numbers. Thus
for any square matrix A the characteristic polynomial/^) factors into linear
factors, f(x) = (x — X^ ri {x — A 2 ) r2 • — (x — X v ) r * where X t ^ A, for i ^j
and r t is the algebraic multiplicity of the eigenvalue X t . The minimum poly
nomial m(x) for A is of the form m(x) = (x — X 1 ) Sl (x — A 2 ) S2 • •  (x — X p ) s v
where 1 < ^ < r^
In the theorems about the diagonalization of matrices we sought bases
made up of eigenvectors. Because we are faced with the possibility that
such bases do not exist, we must seek proper generalizations of the eigen
vectors. It is more fruitful to think of the eigenspaces rather than the
eigenvectors themselves. An eigenvalue is a scalar X for which the linear
transformation a — X is singular. An eigenspace is the kernel (of positive
dimension) of the linear transformation a — X. The proper generalization
of eigenspaces turns out to be the kernels of higher powers of a — X. For a
given eigenvalue X, let A4 fc be the kernel of (a — X) k . Thus, M° = {0} and M 1
8  The Jordan Normal Form 119
is the eigenspace of X. For a e M k , (a  X)* +1 (a) = (a  X){a  Xf(<x) =
(a  Xj(0j= 0. Hence, M fc c M k +K Also, for a e M fc+1 , (a  Xf(a  A) (a) =
(a  A) fc+1 (a) = so that (a  A)(a) 6 M k . Hence, (a  A)/^ 1 c M*.
Since all M* <= y and V is finite dimensional, the sequence of subspaces
M° c M 1 cz M 2 c • • • must eventually stop increasing. Let f be the smallest
index such that M k = M< for all k > t, and denote M* by M (A) . Let m k be the
dimension of M k and w t the dimension of M (A) .
Let ((T  XfV = W k . Then W k + l = (a  X) k+1 V = (a  ^)*{(ct  A)V} <=
(a — X) k V = W k . Thus, the subspaces W k form a decreasing sequence
VV^W 1 ^ W 2 =5 • • • . Since the dimension of W* is « — w fc , we see that
W k = W for all k > t. Denote W* by W M) .
Theorem 8.1. V is the direct sum of M (X) and W (A) .
proof. Since (a  X)W* = (a  Xy+W = W^ 1 = W* we see that a  X
is nonsingular on W* = W (A) . Now let a be any vector in V. Then
(a — A)'(a) = /5 is an element of W (A) . Because (c — A)' is nonsingular on
W (A) there is a unique vector y e W (A) such that (<r — A)'(y) = /?. Let
a — y be denoted by (5. It is easily seen that d e M (A) . Hence V = M (A) + W U) .
Finally, since dim M (A) = w, and dim W (A) = n — m t , the sum is direct. □
In the course of defining M k and W k we have shown that
(1) (a  X)M k +* c M* c M*+i,
(2) (<r  A)W fc = W^ 1 c W*.
This shows that each M k and W fc is invariant under a — X. It follows
immediately that each is invariant under any polynomial in a — X, and
hence also under any polynomial in a. The use we wish to make of this
observation is that if fx is any other eigenvalue, then a — jj, also maps
M (A) and W (A) into themselves.
Let A x , . . . , X v be the distinct eigenvalues of or. Let M { be a simpler
notation for the subspace M (X . } defined as above, and let W, be a simpler
notation for W (A )#
Theorem 8.2. For A t  ^ X jf M< c W..
proof. Suppose a e M,. is in the kernel of a — A,. Then
(/,  A^'a = {(a  A,)  (a  *,)}*(a)
= (a  A,)«(a) +(_1)^W  A,)" fc ((T  A,)*(a).
The first term is zero because a e M i9 and the others are zero because a
is in the kernel of a — A,. Since X j — A, ^ 0, it follows that a = 0. This
means that a — A 3  is nonsingular on M^, that is, a — Xj maps M. onto
120 Determinants, Eigenvalues, and Similarity Transformations  III
itself. Thus M i is contained in the set of images under (a — 1,)'% and hence
M t c W,. □
Theorem 8.3. V = M ± © M 2 © • • • © M p .
proof. Since V = M 2 © VV 2 and M 2 <= W lt we have V = M 2 V^ =
Mi ® {M 2 ® (W 1 n W 2 )}. Continuing in the same fashion, we get V =
M x © • • • © M p © {W l n • • • n W p }. Thus the theorem will follow if
we can show that W = W x n • • • n W p = {0}. By an extension of remarks
already made (<r  A x ) • • • (a — X v ) = q{a) is nonsingular on W; that is,
q(a) maps W onto itself. For arbitrarily large k, [q{o)f also maps W onto
itself. But <7(x) contains each factor of the characteristic polynomial f(x)
so that for large enough k, [q{x)f is divisible by f(x). This implies that
W = {0}. □
Corollary 8.4. t t = Stfor i = 1, . . . , p.
proof. Since V = M x © • • • © M p and (a — W vanishes on M i5 it
follows that (a — X x )^ • • • (a — X v )^ vanishes on all of V. Thus (a; — X x ) tl
• • • (x — Aj,)'" is divisible by the minimum polynomial and s t < t t .
On the other hand, if for a single / we have s t < t t , there is an a e M i
such that (a — A 2 ) Si (a) =^ 0. For all X 5 ^ A t , a — A, is nonsingular on
M { . Hence m(a) ^ 0. This is a contradiction so that t t = s t . D
Let us return to the situation where, for the single eigenvalue X, M k is the
kernel of (a  Xf and W k = (a  X) k V. In view of Corollary 8.4 we let
s be the smallest index such that M k = M s for all k > s. By induction we
can construct a basis {a l9 . . . , a m } of M x such that {a 1? . . . , a TO } is a basis
of M k .
We now proceed step by step to modify this basis. The set {a TO i+1 , . . . ,a m }
consists of those basis elements in M s which are not in M s_1 . These
elements do not have to be replaced, but for consistency of notation we
change their names; let a m$ _ i+v = m§ _ i+ ,. Now set {a  X)(P m _ 1+ ,) =
&._,+, an d consider the set {a l5 . . . , a Ws2 } u {(3 ms _ 2+1 , . . . , P ms _ 2+rils _ ms J.
We wish to show that this set is linearly independent.
If this set were linearly dependent, a nontrivial relation would exist and
it would have to involve at least one of the & with a nonzero coefficient
since the set {a x , . . ., a TOg _J is linearly independent. But then a nontrivial
linear combination of the & would be an element of M $ ~ 2 , and (a — X) s ~ 2
would map this linear combination onto 0. This would mean that (or — A) s_1
would map a nontrivial linear combination of {<x m +1 , . . . , <x m } onto 0.
Then this nontrivial linear combination would be in M s_1 , which would
contradict the linear independence of {a l5 . . . , a }. Thus the set {a x , . . . ,
a ™ 8 _J u {&»,_,+!, • • • » P ms „ 2+ms m s J is linearly independent.
This linearly independent subset of M s_1 can be expanded to a basis of
Ad s_1 We use /Ts to denote these additional elements of this basis, if any
8  The Jordan Normal Form
121
additional elements are required. Thus we have the new basis {a l5 . . . ,
We now set (a — A)(/9 TOs _ 2+ „) = P ms  3 +v anc * P rocee d as before to obtain
a new basis {a l5 . . . , a B( JU {/3 mj _ 3+ V • • • , P m ,J of M s ~ 2 .
Proceeding in this manner we finally get a new basis {&,..., j5 TO } of
M (A) such that {&, . . . , ftj is a basis of M« and (p  X)(P mk+v ) = ?l k _ 1+v
for k > 1 . This relation can be rewritten in the form
a (Pm k +v) ~ W mk+V + P mk _ 1+V
for k > 1,
for v < w v
(8.1)
Thus we see that in a certain sense P mk+V is "almost" an eigenvector.
This suggests reordering the basis vectors so that {/?!, /5 TOi+1 , . . . , /S TOs _ i+1 }
are listed first. Next we should like to list the vectors {/3 2 , p mi+2 , . . .}, etc.
The general idea is to list each of the first elements from each section of the
/Ts, then each of the second elements from each section, and continue until
a new ordering of the basis is obtained.
With the basis of M (A) listed in this order (and assuming for the moment
that that A1 (A) is all of V) the matrix representing a takes the form
s rows
<s rows *
~X 1 • • •
A 1 • • •
I •••
all zeros
all zeros
•• • X
1
• ••
X
X 1 •••
X ■■■
all zeros
••• X
all zeros
i
all zeros
all zeros
etc.
122
Determinants, Eigenvalues, and Similarity Transformations  III
Theorem 8.5. Let A be a matrix with characteristic polynomial f{x) =
(x — A 1 ) ri • ' • (x — 2. p ) r p and minimum polynomial m(x) = (x — A 2 ) Sl • • •
(x — A P ) Sv . A is similar to a matrix J with submatrices of the form
'k
1
K
1
K
••
0"
••
••
•• K
1

K_
B< =
along the main diagonal. All other elements of J are zero. For each X t there
at least one B { of order s<. All other B { corresponding to this A t are of order
less than or equal to s t . The number of B t corresponding to this A f is equal
to the geometric multiplicity of X t . The sum of the orders of all the B t corre
sponding to X t is r t . While the ordering of the B t along the main diagonal
of J is not unique, the number ofB t of each possible order is uniquely determined
by A. J is called the Jordan normal form corresponding to A.
proof. From Theorem 8.3 we have V = M x e • • • M„. In the dis
cussion preceding the statement of Theorem 8.5 we have shown that each
M; has a basis of a special type. Since V is the sum of the M if the union of
these bases spans V. Since the sum is direct, the union of these bases is
linearly independent and, hence, a basis for V. This shows that a matrix
J of the type described in Theorem 8.5 does represent a and is therefore
similar to A.
The discussion preceding the statement of the theorem also shows that
the dimensions m ik of the kernels Mf of the various (a — X t ) k determine
the orders of the B t in /. Since A determines a and a determines the subspace
M 4 * independently of the bases employed, the B t are uniquely determined.
Since the X i appear along the main diagonal of / and all other nonzero
elements of / are above the main diagonal, the number of times x — A t
appears as a factor of the characteristic polynomial of /is equal to the number
of times X t appears in the main diagonal. Thus the sum of the orders of the
Bi corresponding to 2. { is exactly r t . This establishes all the statements of
Theorem 8.5. □
Let us illustrate the workings of the theorems of this section with some
examples. Unfortunately, it is a little difficult to construct an interesting
8 I The Jordan Normal Form
123
example of low order. Hence, we give two examples, The first example
illustrates the choice of basis as described for the space M {X) . The second
example illustrates the situation described by Theorem 8.3.
Example 1. Let
~ 1 01 1
0~
4 13 2
1
A =
21 1
1
3 1 3 4
1
_8 2 7 5
4_
The first step is to obtain the characteristic matrix
~\x 1 1
o 
4 1 a; 3 2
1
C{x) =
2 1 x 1
1
3 1 3 4  x
1
_ ~ 8
2 7 5
4
— x
Although it is tedious work we can obtain the characteristic polynomial
fix) = (x — 2) 5 . We have one eigenvalue with algebraic multiplicity 5.
What is the geometric multiplicity and what is the minimum equation for
A1 Although there is an effective method for determining the minimum
equation, it is less work and less wasted effort to proceed directly with
determining the eigenvectors. Thus, from
1 01 1 0"
4 13 2 1
C(2) = 2 1 2 1 1
3 13 2 1
8 2 7 5 2_
we obtain by elementary row operations the Hermite normal form
1
1
124 Determinants, Eigenvalues, and Similarity Transformations  III
From this we learn that there are two linearly independent eigenvectors
corresponding to 2. The dimension of M 1 is 2. Without difficulty we find
the eigenvectors
a a = (0,1,1,1,0)
a 2 = (0, 1,0,0, 1).
Now we must compute (A — 21 f = (C(2)) 2 , and obtain
0"
{A  2/) 2 =
1 01 1
1 01 1 o
1
1
1
The rank of {A  2/) 2 is 1 and hence M 2 is of dimension 4. The a x and <x 2
we already have are in M 2 and we must obtain two more vectors in M 2
which, together with a x and a 2 , will form an independent set. There is
quite a bit of freedom for choice and
a 3 = (0,1,0,0,0)
a 4 = (1,0,1,0,0)
appear to be as good as any.
Now {A — 2/) 3 = 0, and we know that the minimum polynomial for A
is (x — 2) 3 . We have this knowledge and quite a bit more more for less work
than would be required to find the minimum polynomial directly. We see,
then, that M 3 = V and we have to find another vector independent of a l5
a 2 , a 3 , and a 4 . Again, there are many possible choices. Some choices will
lead to a simpler matrix of transition than will others, and there seems to
be no very good way to make the choice that will result in the simplest
matrix of transition. Let us take
<x 6 = (0,0,0,1,0).
We now have the basis of {a,, a 2 , a 3 , a 4 , a 5 } such that {a 1? a 2 } is a basis
of M 1 , {a x , a 2 , a 3 , a 4 } is a basis of M 2 , and {a l5 a 2 , a 3 , a 4 , a 5 } is a basis of
M 3 . Following our instructions, we set /? 5 = <x 5 . Then
(A
21)
~0~
~1~
2
=
1
1
*
_0_
_5_
8 j The Jordan Normal Form
125
Hence, we set fa = (1, 2, 1, 2, 5). Now we must choose fa so that {a l3
a 2> fa> &} is a basis for M 2 . We can choose fa = (—1,0, 1,0, 0). Then
(A  21)
~r
~0~
{A  21)
~r
~o
2
i
1
=
1
i
=
2
1
_5_
_1_
^
o_
1
Hence, we choose fa = (0, 0, 1, 1, 1) and fa = (0, 1, 0, 0, 1). Thus,
P =
"0
1
1"
2
1
1
1
1
1
2
1
1
5
1
is the matrix of transition that will transform A to the Jordan normal form
Example 2. Let
A =
"2
1
0~
2
1
=
2
2
1
_0
2_
"5
1
3
2
5"
2
1
1
1
2
1
3
1
1
1
1
1
1
The characteristic polynomial is f{x) = —(x — 2) 3 (x — 3) 2 . Again we have
126 Determinants, Eigenvalues, and Similarity Transformations  HI
repeated eigenvalues, one of multiplicity 3 and one of multiplicity 2.
C(2) =
01
1 1 1
from which we obtain the Hermite normal form
"3
1
3
1
1
2
5~
1
2
1
1
1
1_
1
1
1
2~
1
1
0_
Again, the geometric multiplicity is less than the algebraic multiplicity. We
obtain the eigenvectors
a x = (1,0,1,0,0)
a 2 = (2,1,0,0,1).
Now we must compute (A — 2/) 2 . We find
~1
1
2"
(A  2/) 2 =
1
2
1
2
1
1
1
1
1
from which we obtain the Hermite normal form
~l 01 2"
1
1
1
8 I The Jordan Normal Form
127
For the third basis vector we can choose
a 3 = (0, 1,0, 1,0).
Then
(A  21)
~0~
~1~
1
=
1
1
_0_
_0_
hence, we have /3 3 = a 3 , /3 X = a 1( and we can choose /? 2 = a 2 .
In a similar fashion we find & = (1, 0, 0, 1, 0) and & = (2, 0, 0, 0, 1)
corresponding to the eigenvalue 3. /? 4 is an eigenvector and {A — 3I)fi 5 = S 4 .
chapter
IV Linear
functionals,
bilinear forms,
quadratic forms
In this chapter we study scalar valued functions of vectors. Linear functional
are linear transformations of a vector space into a vector space of dimension
1. As such they are not new to us. But because they are very important, they
have been the subject of much investigation and a great deal of special
terminology has accumulated for them.
For the first time we make use of the fact that the set of linear transforma
tions can profitably be considered to be a vector space. For finite dimensional
vector spaces the set of linear functionals forms a vector space of the same
dimension, the dual space. We are concerned with the relations between
the structure of a vector space and its dual space, and between the representa
tions of the various objects in these spaces.
In Chapter V we carry the vector point of view of linear functionals one
step further by mapping them into the original vector space. There is a
certain aesthetic appeal in imposing two separate structures on a single
vector space, and there is value in doing it because it motivates our con
centration on the aspects of these two structures that either look alike or
are symmetric. For clarity in this chapter, however, we keep these two
structures separate in two different vector spaces.
Bilinear forms are functions of two vector variables which are linear in
each variable separately. A quadratic form is a function of a single vector
variable which is obtained by identifying the two variables in a bilinear
form. Bilinear forms and quadratic forms are intimately tied together,
and this is the principal reason for our treating bilinear forms in detail.
In Chapter VI we give some applications of quadratic forms to physical
problems.
If the field of scalars is the field of complex numbers, then the applications
128
1 I Linear Functionals 129
we wish to make of bilinear forms and quadratic forms leads us to modify
the definition slightly. In this way we are led to study Hermitian forms.
Aside from their definition they present little additional difficulty.
1 I Linear Functionals
Definition. Let V be a vector space over a field of constants F. A linear
transformation cf> of V into F is called a linear form or linear functional on V.
Any field can be considered to be a 1 dimensional vector space over itself
(see Exercise 10, Section 11). It is possible, for example, to imagine two
copies of F, one of which we label U. We retain the operation of addition in
U, but drop the operation of multiplication. We then define scalar multi
plication in the obvious way : the product is computed as if both the scalar
and the vector were in the same copy of F and the product taken to be an
element of U. Thus the concept of a linear functional is not really something
new. It is our familiar linear transformation restricted to a special case.
Linear functionals are so useful, however, that they deserve a special name
and particular study. Linear concepts appear throughout mathematics
particularly in applied mathematics, and in all cases linear functionals play
an important part. It is usually the case, however, that special terminology
is used which tends to obscure the widespread occurrence of this concept.
The term "linear form" would be more consistent with other usage
throughout this book and the history of the theory of matrices. But the
term "linear functional" has come to be almost universally adopted.
Theorem 1.1. If V is a vector space of dimension n over F, the set of all
linear functionals on V is a vector space of dimension n.
proof. If <f> and ip are linear functionals on V, by <f> + tp we mean the
mapping defined by (<f> + ^)(a) = <£(a) + y>(a) for all a e V. For any
a e F, by a<f> we mean the mapping defined by (a<£)(a) = a [<£(<*)] for all
a e V. We must then show that with these laws for vector addition and
scalar multiplication of linear functionals the axioms of a vector space are
satisfied.
These demonstrations are not difficult and they are left to the reader.
(Remember that proving axioms Al and B\ are satisfied really requires
showing that <j> + y> and a<j>, as defined, are linear functionals.)
We call the vector space of all linear functionals on V the dual or conjugate
space of V and denote it by V (pronounced "vee hat" or "vee caret"). We have
yet to show that Vis of dimension n. Let A = {a l5 a 2 , . . . , a w } be a basis of V.
Define fa by the rule that for any a = JjLi **<**, &( a ) = a i e F  We sha11
call <f>i the ith coordinate function.
130 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
For any /8 = £Li b^ we have <f> t (P) = b, and &(a + 0) = &2JU ^ +
2?=i V*> = <MIti («* + ^K> = a< + ^ = &(a) + &(£)• Also &(aa) =
&{ fl X"=i a y a /) = &Gt£=i «^ a ^} = «o» = ^t(a) Thus <^ is a linear func
tional.
Suppose that 2» =1 Z>^. = 0. Then Q> =1 ^)(a) = for all a e V.
In particular for a< we have (2" =1 ^XaJ = 2*=xM*( a <) = *< = ° Hence,
all Z) t = and the set {<{>!, <f> 2 , ■ ■ ■ , <f> n ) must be linearly independent. On the
other hand, for any <£ e V and any a = JjL x «*<** e V, we have
In \ n
If we let ^(aj = 6 if then for ^J=i ^</v we have
(iM/W) = 2 MX") = 2 My = #«)• (L2)
\y=i / y=i y=i
A A A.
Thus the set {<f> x , . . . , n } = A spans V and forms a basis of V. This shows
that V is of dimension n. □
A. A.
The basis A of V that we have constructed in the proof of Theorem 1.1
has a very special relation to the basis A. This relation is characterized by
the equations
4>iM = tii> (13)
for all i, j. In the proof of Theorem 1 . 1 we have shown that a basis satisfying
these conditions exists. For each i, the conditions in Equation (1.3) specify
the values of ^> t on all the vectors in the basis A. Thus </>; is uniquely deter
mined as a linear functional. And thus A is uniquely determined by A and the
conditions (1.3). We call A the basis dual to the basis A.
n
<K*t) = 2 bdlcLj) = b,
i=l
so that, as a linear transformation, <f> is represented by the 1 x n matrix
\b x ' * * b n ]. For this reason we choose to represent the linear functionals in
A A A
V by onerow matrices. With respect to the basis A in V, <f> = 2?=i b^i w iU
be represented by the row [b x • • • b n ] = B. It might be argued that, since V
is a vector space, the elements of V should be represented by columns. But
the set of all linear transformations of one vector space into another also
forms a vector space, and we can as justifiably choose to emphasize the aspect
of V as a set of linear transformations. At most, the choice of a representing
notation is a matter of taste and convenience. The choice we have made
means that some adjustments will have to be made when using the matrix
1 I Linear Functionals
131
of transition to change the coordinates of a linear functional when the basis
is changed. But no choice of representing notation seems to avoid all such
difficulties and the choice we have made seems to offer the most advantages.
If the vector £ e V is represented by the ntuple (x lt . . . , x n ) = X, then
we can compute <£(£) directly in terms of the representations.
n n
3 = 1
= [b t ■■•b n ]
= BX.
(1.4)
EXERCISES
1. Let A = {a l5 <x 2 , a 3 } be a basis in a 3dimensional vector space V over R.
Let A ={<f> x , <f> 2 , <f> 3 } be the basis in V dual to A. Any vector I G V can be written in
the form £ = x 1 a 1 + z 2 a 2 + x a a. 3 . Determine which of the following functions
on V are linear functionals. Determine the coordinates of those that are linear
.A.
functionals in terms of the basis A.
(a) <£(£) = *i + x 2 + x z
(b) $(!;) =(x 1 _ + x 2 )\
(c) <£(!) = ^2x v
{d) <£(£) =x 2  %x v
(e) m =x 2 l
A
2. For each of the following bases of R 3 determine the dual basis in R 3 .
(a) {(1,0,0), (0,1,0), (0,0, 1)}.
(b) {(1,0,0), (1,1,0), (1,1,1)}.
(c) {(1,0, 1), (1,1,0), (0,1,1)}.
3. Let V = P n , the space of polynomials of degree less than n over R. For a
fixed aeR, let </>(/?) = p ik) (a), where p ( ^(x) is the fcth derivative of p(x)eP n .
Show that <f> is a linear functional.
4. Let V be the space of real functions continuous on the interval [0, 1], and let
g be a fixed function in V. For each/e V define
Jo
fWOdt.
132 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
Show that L g is a linear functional on V. Show that if L g {f) = for every geV,
then/ = 0.
^ 5. Let A = {04, . . . , a n } be a basis of V and let A = {^, . . . , <f> n } be the basis of
V dual to the basis A. Show that an arbitrary a g V can be represented in the form
n
a = 2 &( a ) a i
6. Let V be a vector space of finite dimension n > 2 over F. Let a and /S be two
vectors in V such that {a, p) is linearly independent. Show that there exists a
linear functional <f> such that <£(<*) = 1 and <f>(P) = 0.
7. Let V = P n , the space of polynomials over F of degree less than n(n > 1). Let
a g F be any scalar. For each p(x) e P n , p(a) is a scalar. Show that the mapping
of p(x) onto p(a) is a linear functional on P„ (which we denote by a a ). Show that if
a 7^ b then <r a ^ cr b .
8. (Continuation) In Exercise 7 we showed that for each ae F there is defined
A.
a linear functional <r a e P n . Show that if « > 1 , then not every linear functional in
P n can be obtained in this way.
9. (Continuation) Let {a x , . . . , a n } be a set of n distinct scalars. Let/0») =
(x  a x ){x  a 2 ) • • • (x  a n ) and h k (x) =f(x) =f(x)/(x  a k ). Show that h k {a 3 ) =
^ikf'i^j), where/'(#) is the derivative off(x).
10. (Continuation) For the a k given in Exercise 9, let
1
3 /'K> °'"
Show that {ct x , . . . , a n } is linearly independent and a basis of P n . Show that
{h x {x), . . . , /?„(x)} is linearly independent and, hence, a basis of P n . (Hint: Apply
a j to 2fc=i ^fc^/cC^)) Show that {ffj, . . . , a n ) is the basis dual to {h^ix), . . . , /? n (a;)}
11. (Continuation) Let p(x) be any polynomial in P n . Show that p(x) can be
represented in the form
« /?(a fc )
*;=i / ^^
(Hint: Use Exercise 5.) This formula is known as the Lagrange interpolation
formula. It yields the polynomial of least degree taking on the n specified values
{p(a 1 ), . . . , p(a n )} at the points {a lf . . . , a n ).
12. Let W be a proper subspace of the »dimensional vector space V. Let a be
a vector in V but not in W. Show that there is a linear functional <£ G y such that
<A(a ) = 1 and 0(a) = for all a g W.
13. Let W be a proper subspace of the Mdimensional vector space V. Let y>
be a linear functional on W. It must be emphasized that y> is an element of W
2  Duality 133
a a
and not an element of V. Show that there is at least one element <f> e V such that
<f> coincides with y> on W.
14. Show that if a ^ 0, there is a linear functional <f> such that <f>(a) ^ 0.
15. Let a and /S be vectors such that <£(/?) = implies <f>(a) = 0. Show that a is
a multiple of /3.
2 I Duality
Until now, we have encouraged an unsymmetric point of view with respect
to V and V. Indeed, it is natural to consider <£(<x) for a chosen <f> and a range
of choices for a. However, there is no reason why we should not choose a
fixed a and consider the expression <£(a) for a range of choices for <f>. Since
iPi^i + ^2^2)( a ) = (£i<£i)(a) + (6 2 <£ 2 (a), we see that a behaves like a linear
functional on V.
This leads us to consider the space V of all linear functional on V. Corre
sponding to any a £ V we can define a linear functional a in V by the rule
a(<£) = <j>((x.) for all <f> £ V. Let the mapping defined by this rule be denoted
by J, that is, 7(a) = a. Since /(aa + bfi)(<f>) = <f>(a<x. + fyff) = fl^(a) +
60(0) = aJ(a)(<f>) + &/(j8)(0) = [aJ(a) + bJ(p)](cf>) we see that / is a linear
transformation mapping V into 9.
Theorem 2.1. If V is finite dimensional, the mapping J of V into V is a
onetoone linear transformation of V onto V.
proof. Let V be of dimension n. We have already shown that J is linear
and into. If 7(a) = then /(a)(0) = for all <f> £ V. In particular, 7(a)(^) =
for the basis of coordinate functions. Thus if a = 2X X a i x i we see tnat
J(a)(6) = &(a) = 5>,6(a,) = a, =
for each / = 1, . . . , n. Thus a = and the kernel of / is {0}, that is, J{V)
A it
is of dimension n. On the other hand, if V is of dimension n, then V and V
are also of dimension n. Hence J(V) = V and the mapping is onto. □
If the mapping J of V into V is actually onto V we say that V is reflexive.
Thus Theorem 2.1 says that a finite dimensional vector space is reflexive.
Infinite dimensional vector spaces are not reflexive, but a proof of this
assertion is beyond the scope of this book. Moreover, infinite dimensional
vector spaces of interest have a topological structure in addition to the
algebraic structure we are studying. This additional condition requires
a more restricted definition of a linear functional. With this restriction
134 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
the dual space is smaller than our definition permits. Under these condition
it is again possible for the dual of the dual to be covered by the mapping J.
Since J is onto, we identify V and J(V), and consider V as the space of
linear functionals on V. Thus V and V are considered in a symmetrical
position and we speak of them as dual spaces. We also drop the parentheses
from the notation, except when required for grouping, and write </><x instead
of </>(<x). The bases {a. x , . . . , a. n } and {(/>!,..., <f> n } are dual bases if and only
if ^a, = 6 it .
EXERCISES
1. Let A = {oL lf . . . , a„} be a basis of V, and let A = {<f> lt . . . , <£ n } be the basis of
A A.
V dual to the basis A. Show that an arbitrary <f>eV can be represented in the form
n
2. Let V be a vector space of finite dimension n > 2 over F. Let </> and y> be two
linear functionals in V such that {<f>, y>} is linearly independent. Show that there
exists a vector a such that <£(a) = 1 and y(a) = 0.
3. Let <f> be a linear functional not in the subspace S of the space of linear
functionals V. Show that there exists a vector a such that <£ ( a ) = 1 and <£(<x) =0
for all <f>eS.
4. Show that if <f> ¥> 0, there is a vector a such that <£(a) ^ 0.
5. Let <f> and y be two linear functionals such that <f>{a) = implies tp(cc) = 0.
Show that v is a multiple of <f>.
3 I Change of Basis
If the basis A' = {04, x' 2 , . . . , a'J is used instead of the basis A =
{a l5 a 2 , . . . , a n }, we ask how the dual basis A' = {<f>[, . . . , <f>' n } is related to
the dual basis A = {</> l5 . . . , cf> n }. Let i> = [p {j ] be the matrix of transition
from the basis A to the basis A'. Thus a J = ^Li/?^. Since </>j(a 3 ') =
SU/VMO =/>« we see that ^= ^xPa^r A This means that P T is the
matrix of transition from the basis A' to the basis A. Hence, (P 2 ') 1 = (P~ X ) T
A A
is the matrix of transition from A to A'.
Since linear functionals are represented by row matrices instead of column
matrices, the matrix of transition appears in the formulas for change of
coordinates in a slightly different way. Let B = [b x ■ • • b n ] be the representa
tion of a linear functional <f> with respect to the basis A and B' = [b[ • • • b' n ]
3  Change of Basis 135
be its representation with respect to the basis A'. Then
n in
=i(i^A^; (3D
Thus,
B' = BP. (3.2)
We are looking at linear functionals from two different points of view.
Considered as a linear transformation, the effect of a change of coordinates
is given by formula (4.5) of Chapter II, which is identical with (3.2) above.
Considered as a vector, the effect of a change of coordinates is given by
formula (4.3) of Chapter II. In this case we would represent 4> by B T , since
vectors are represented by column matrices. Then, since (P _1 ) r is the
matrix of transition, we would have
b t = (pyB ,T = (B'py,
or (3.3)
B = B'P~\
which is equivalent to (3.2). Thus the end result is the same from either
point of view. It is this twosided aspect of linear functionals which has
made them so important and their study so fruitful.
Example 1. In analytic geometry, a hyperplane passing through the
origin is the set of all points with coordinates (x x , x 2 , . . . , x n ) satisfying an
equation of the form b^ + b 2 x 2 + • • • + b n x n = 0. Thus the «tuple
[b x b 2 • • • b n ] can be considered as representing the hyperplane. Of course,
a given hyperplane can be represented by a family of equations, so that
there is not a onetoone correspondence between the hyperplanes through
the origin and the wtuples. However, we can still profitably consider the
space of hyperplanes as dual to the space of points.
Suppose the coordinate system is changed so that points now have the
coordinates (y l5 . . . , y n ) where x t = 2" =i aaVy Then the equation of the
hyperplane becomes
n n
 n

2 b i x i = 1 b i
^aaVi
i=l i=l
_3' =1
n
 n
= 1
2 b i a a
y,
i=1
_*'=!
n
= 2^, = o.
j=l
(3.4)
136 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
Thus the equation of the hyperplane is transformed by the rule c, = 2*=i ^i a a
Notice that while we have expressed the old coordinates in terms of the new
coordinates we have expressed the new coefficients in terms of the old
coefficients. This is typical of related transformations in dual spaces.
Example 2. A much more illuminating example occurs in the calculus of
functions of several variables. Suppose that w is a function of the variables
#!, x 2 , . . . , x n , w =f(x x , x 2 , . . . , x n ). Then it is customary to write down
formulas of the following form:
, dw , dw , . 9w , '
dw = ax x H dx 2 + • • • \ dx n , (3.5)
dx x dx 2 dx n
and
\dxx ' dx 2 ' ' dxj
dw is usually called the differential of w, and Viv is usually called the gradient
of w. It is also customary to call Viv a vector and to regard dw as a scalar,
approximately a small increment in the value of w.
The difficulty in regarding Vw as a vector is that its coordinates do not
follow the rules for a change of coordinates of a vector. For example, let
us consider (x x , x 2 , . . . , x n ) as the coordinates of a vector in a linear vector
space. This implies the existence of a basis {a 1? . . . , a n } such that the linear
combination
I = 2 ^ (37)
is the vector with coordinates (x x , x 2 , . . . , x n ). Let {fix, . . . , /?„} be a new
basis with matrix of transition P = [p i} ] where
n
Pi = lPifi< (38)
Then, if £ = 2" =1 2/*& is the representation of  in the new coordinate system,
we would have
n
x i=^ViiV^ (39)
or
*« = i:r^ (310)
Let us contrast this with the formulas for changing the coordinates of Vw.
From the calculus of functions of several variables we know that
dw ~ dw dx i
— =2, • (3H)
dy, i=idx { dy 5
3  Change of Basis I37
This formula corresponds to (3.2). Thus Vw changes coordinates as if it were
in the dual space.
In vector analysis it is customary to call a vector whose coordinates change
according to formula (3.10) a contravariant vector, and a vector whose
coordinates change according to formula (3.11) a covariant vector. The
reader should verify that if P =
~dx l
Thus (3.11) is equivalent to the formula
then
'dy t '
dx 4 _
= (P T )~\ Thus
dw ^ dy* dw
r=2rT' 3  12 )
From the point of view of linear vector spaces it is a mistake to regard
both types of vectors as being in the same vector space. As a matter of fact,
their sum is not defined. It is clearer and more fruitful to consider the co
variant and contravariant vectors to be taken from a pair of dual spaces.
This point of view is now taken in modern treatments of advanced calculus
and vector analysis. Further details in developing this point of view are given
in Chapter VI, Section 4.
In traditional discussions of these topics, all quantities that are represented
by ntuples are called vectors.
In fact, the «tuples themselves are called vectors. Also, it is customary
to restrict the discussion to coordinate changes in which both covariant
and contravariant vectors transform according to the same formulas. This
amounts to having P, the matrix of transition, satisfy the condition (P X ) r =
P. While this does simplify the discussion it makes it almost impossible to
understand the foundations of the subject.
Let A = {«!, . . . , a J be a basis of V and let A = {<f> x , ...,<£ J be the dual
basis in V. Let 8 = {fi lt . . . , pj be any new basis of V. We are asked to find
the dual basis /S in V. This problem is ordinarily posed by giving the repre
sentation of the fa with respect to the basis A and expecting the representations
of the elements of the dual basis with respect of A. Let the fa be represented
with respect to A in the form
n
Pi = I,Pii<Xi> (3.13)
and let
*Pi=2qif<f>i (3.14)
3 = 1
be the representations of the elements of the dual bases B = {y^, . . . , y>J.
138 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
Then
<5*i = Vtft = ( 2 «*<&)( %Pn*\
n n
i=l j=l
n
= J f 1kiPa (315)
i=l
In matrix form, (3.15) is equivalent to
I=QP. (3.16)
(2 is the inverse of P. Because of (3.15), the ip t are represented by the rows
of Q. Thus, to find the dual basis, we write the representation of the basis 8
in the columns of P, find the inverse matrix P _1 , and read out the repre
A.
sentations of the basis 6 in the rows of P~ x .
EXERCISES
1. Let A = {(1, 0, . . . , 0), (0, 1, . . . , 0), . . . , (0, 0, . . . , 1)} be a basis of R n .
The basis of R n dual to A has the same coordinates. It is of interest to see if there
are other bases of R n for which the dual basis has excatly the same coordinates.
Let A' be another basis of R n with matrix of transition P. What condition should
P satisfy in order that the elements of the basis dual to A' have the same coordinates
as the corresponding elements of the basis A' ?
2. Let A = {a l9 a 2 , a 3 } be a basis of a 3dimensional vector space V, and let A =
{<t>i, K <&}} be the basis of V dual to A. Then let A' ={(1,1, 1), (1, 0, 1), (0, 1, 1)}
be another basis of V (where the coordinates are given in terms of the basis A).
Use the matrix of transition to find the basis A' dual to A'.
3. Use the matrix of transition to find the basis dual to {(1,0,0), (1, 1,0),
(1,1,1)}.
4. Use the matrix of transition to find the basis dual to {(1, 0, —1), ( — 1, 1, 0),
(0,1,1)}.
5. Let B represent a linear functional $, and X a vector £ with respect to dual
bases, so that BX is the value <f>$ of the linear functional. Let P be the matrix of
transition to a new basis so that if X' is the new representation of I, then X = PX'.
By substituting PX' for X in the expression for the value of <f>£ obtain another
proof that BP is the representation of <f> in the new dual coordinate system.
4 I Annihilators
A.
Definition. Let V be an wdimensional vector space and V its dual. If, for
A.
an a g V and a <f> e V, we have <f>a. = 0, we say that <f> and a are orthogonal.
4 I Annihilators
139
Since <f> and a are from different vector spaces, it should be clear that we do
not intend to say that the <f> and a are at "right angles."
Definition. Let W be a subset (not necessarily a subspace) of V. The set of
all linear functional <f> such that <£a = for all a e W is called the annihilator
of W, and we denote it by V\A. Any <f> e V\A is called «« annihilator of W.
Theorem 4.1. The annihilator W x o/ W is a subspace of V. If W is a
subspace of dimension p, then W 1 is of dimension n — p.
proof. If <f> and tp are in W 1 , then (a^ + by>)a. = a <f>* + biptx. = for
all a e W. Hence, W L is a subspace of V.
Suppose W is a subspace of V of dimension p, and let A = {a lf . . . , a B }
be a basis of V such that {a l5 . . . , a p } is a basis of W. Let A = {fa, . . . , <£J
be the dual basis of A. For {<f> p+1 , . . . , fa} we see that ^a, = fo'r all
i < p. Hence, {0 lH . 1 , . . . , fa} is a subset of the annihilator of W. On the
other hand, if <j> = £ n =1 brf> t is an annihilator of W, we have 0<x, = for
each / < p. But fa = J; =1 ^ a , = 6,. Hence, b t = for i ^ p and the
set {fa +1 , . . . , fa} spans W^. Thus {<£ p+1 , . . . , fa} is a basis for W^,
and W x is of dimension n  p. The dimension of W^ is called the co
dimension of W. □
It should also be clear from this argument that W is exactly the set of all
a e V annihilated by all <£ e W 1 . Thus we have
Theorem 4.2. If S is any subset of V, the set of all a. e V annihilated by all
<f>eSis a subspace of V, denoted by S±. If S is a subspace of dimension r,
then S 1 is a subspace of dimension n — r. □
Theorem 4.2 is really Theorem 1.16 of Chapter II in a different form. If
a linear transformation of V into another vector space W is represented by
a matrix A, then each row of A can be considered as representing a linear
functional on V. The number r of linearly independent rows of A is the
dimension of the subspace S of V spanned by these linear functional. S 1
is the kernel of the linear transformation and its dimension is n — r.
The symmetry in this discussion should be apparent. If <f> e W 1 , then
<f>a. = for all a e W. On the other hand, for a e W, <£a = for all (f>eW^.
Theorem 4.3. If W is a subspace, (W^ = W.
proof. By definition, (W^ = W±± is the set of a e V such that
<A<x = for all <f> e W^. Clearly, W c W 11 . Since dim W 11 = n 
dim Wl = dim W, W ±x = W. □
This also leads to a reinterpretation of the discussion in Section II8.
A subspace W of V of dimension p can be characterized by giving its
annihilator W 1  c V of dimension /• = n ~ p.
140 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
Theorem 4.4. If W x and W 2 are two subspaces of V, and VV 1 and VV 1
are their respective annihilators in V, the annihilator of W 1 + VV 2 is VV 1 n
W 2 ± and the annihilator of W x n VV 2 is W 1 + VV 1 .
proof. If <f> is an annihilator of VV X + W l5 then <f> annihilates all a g W x
and all g W 2 so that </> g VV 1 n VV 1 . If e VV 1 n VV 1 , then for all
a g VV X and /S g VV 2 we have </><x = and </>/? = 0. Hence, (f>(aa. + 6/?) =
a<£oc + b<f>p = so that annihilates W^ + VV 2 . This shows that (W x +
w,) 1 = VV 1 n VV 1 .
The symmetry between the annihilator and the annihilated means that
the second part of the theorem follows immediately from the first. Namely,
since (W x + VVj) 1 = VV 1 n VV 1 , we have by substituting VV 1 and W 1
for W x and W 2 , (VV 1 + VV 1 ) 1 = (VV 1 ) 1 n (VV 1 ) 1 = VV X n VV>. Hence,
(W 1 n VV,) 1 = VV 1 + VV 1 . n
Now the mechanics for finding the sum of two subspaces is somewhat
simpler than that for finding the intersection. To find the sum we merely
combine the two bases for the two subspaces and then discard dependent
vectors until an independent spanning set for the sum remains. It happens
that to find the intersection W x n W 2 it is easier to find VV 1 and W 1
and then VV 1 + VV 1 and obtain W x n W 2 as (VV 1 + VV 1 ) 1 , than it is to
find the intersection directly.
The example in Chapter II8, page 71, is exactly this process carried out
in detail. In the notation of this discussion £ x = W 1 and £ 2 = VV 1 .
A
Let V be a vector space, V the corresponding dual vector space, and let VV
be a subspace of V. Since VV <= V, is there any simple relation between VV
and V? There is a relation but it is fairly sophisticated. Any function defined
on all of V is certainly defined on any subset. A linear functional <f> e V,
therefore, defines a function on VV, which we have called the restriction of <j>
A A
to VV. This does not mean that V <= VV; it means that the restriction defines
A A
a mapping of V into VV.
Let us denote the restriction of </> to VV by <j>, and denote the mapping of <f>
onto 4> by R. We call R the restriction mapping. It is easily seen that R is
linear. The kernel of R is the set of all <f> g V such that </>(a) = for all
a g VV. Thus K(R) = VV 1 . Since dim VV = dim VV = n  dim VV 1 =
n — dim K(R), the restriction map is an epimorphism. Every linear functional
on VV is the restriction of a linear functional on V.
A A
Since K(R) = VV 1 , we have also shown that VV and V/VV 1 are isomorphic.
But two vector spaces of the same dimension are isomorphic in many ways.
We have done more than show that VV and V/VV 1 are isomorphic. We have
shown that there is a canonical isomorphism that can be specified in a natural
way independent of any coordinate system. If <f> is a residue class in V/W 1 ,
4  Annihilators j4j
and <f> is any element of this residue class, then $ and R(<f>) correspond under
this natural isomorphism. If rj denotes the natural homomorphism of
V onto V/W 1 , and t denotes the mapping of 4> onto R(<f>) defined above,
then R = Tt], and t is uniquely determined by R and t] and this relation.
Theorem 4.5.^ Let W be a subspace of V and let W 1 be the annihilator of
W in V.^ ThenW is isomorphic to V/W 1 . Furthermore, if R is the restriction
map of \ onto W, ifrj is the natural homomorphism of V onto V/W 1 , and r is the
unique isomorphism ofV/W^ onto W characterized by the condition R = rrj,
then r{<f>) = R(cf>) where <j> is any linear functional in the residue class $ e
V/W ± . a
EXERCISES
1 . (a) Find a basis for the annihilator of W = <(1 , 0,  1), (1 ,  1 , 0), (0, 1 ,  1)>.
(b) Find a basis for the annihilator of W = <(1, 1, 1, 1,1), '(1, 0, 1,0, 1)' (0 1
1,1,0), (2, 0, 0, 1, 1), (2, 1,1,2, 1), (1,1,1, 2, 2), (1, 2, 3, 4, 1)>. What
are the dimensions of W and W^ ?
2. Find a nonzero linear functional which takes on the same nonzero value
for ^ = (1, 2, 3), £ 2 = (2, 1, 1), and f 8 = (1, 0, 1).
3. Use an argument based on the dimension of the annihilator to show that if
a * 0, there is a <f> e V such that </>«. # 0.
4. Show that if S c 7", then S x = 7^.
5. Show that <S> = S^ 1 .
6. Show that if S and T are subsets of V each containing 0, then
(S + T)L <= SJ n TL,
and
S 1 + T 1 c (S n T)L.
7. Show that if S and T are subspaces of V, then
(S + 7> = S^ n TL,
and
S 1 + TL = (S n T)L.
8. Show that ifS and T are subspaces of V such that the sum S + T is direct,
then S 1  + TL = y.
9. Show that if S and T are subspaces of V such that S + T = V, then SL n
T± = {0}.
10. Show that if S and T are subspaces of V such that S ®T = V, then V =
SL TL. Show that SL is isomorphic to T and that TL is isomorphic to S.
11. Let V be vector space over the real numbers, and let tf> be a nonzero linear
functional on V. We refer to the subspace S of V annihilated by ^ as a hyperplane
of V. Let S+ = {a  0(«) > 0}, and S~ = {a  0(a) < 0}. We call S+ and S~ the two
142 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
sides of the hyperplane S. If a and P are two vectors, the line segment joining a
and (3 is defined to be the set {to. + (1 — t)fi  <, t < 1}, which we denote by
a/?. Show that if a and /? are both in the same side of S, then each vector in a/S is
also in the same side. And show that if a and /? are in opposite sides of S, then a/?
contains a vector in S.
5 I The Dual of a Linear Transformation
Let U and V be vector spaces and let a be a linear transformation mapping
U into V. Let V be the dual space of V and let be a linear functional on V.
For each «eU, (r(<x) g V so that can be applied to (r(a). Thus (/>[cr(a)] £ F
and (f>a can be considered to be a mapping which maps U into F. For
a, 8 e 1/ and a,b eF we have <£[(r(aa + 6/?)] = </>>(7(a) + 6<x(#)] =
a^cr(a) + bcf)o({3) so that we have shown
A.
Theorem 5.1. For a a linear transformation of U into V, and cf> eV, the
mapping <f>a defined by <f>[c(<x)] = <f>a((x) is a linear functional on U; that is,
(f>a eil. □
Theorem 5.2. For a given linear transformation a mapping U into V, the
mapping of V into U defined by making <j> e V correspond to <f>o g U is a linear
transformation of V into U.
proof. For l5 4> 2 g V and a, b e F, {a<f> x + b(f> 2 )o(<x) — A0 1 cr(a) + txj> 2 a{cf)
for all a e U so that a§\ + b(f> 2 in V is mapped onto a<f> r a + b(f> 2 o e U and
the mapping defined is linear. □
A. A.
Definition. The mapping of c/> e V onto <f>o g U is called the dual of a and
is denoted by a. Thus a{<f>) — <f>a.
Let A be the matrix representing a with respect to the bases A in U and 8
A A. A A
in V. Let A and B be the dual bases in U and V, respectively. The question
now arises: "How is the matrix representing a with respect to the bases
A A
8 and A related to the matrix representing a with respect to the bases A and 8 ?"
For A = {a l5 . . . , a TO } and 8 = {fi lt . . . , n } we have <r(a,) = 2f =1 a w ft.
Let {A, . . . , (£ TO } be the basis of U dual to A and let {y) x , . . . , y n ) be the basis
A A
of V dual to 8. Then for ^sVwe have
= Vi( Jxa)
(5.1)
5  The Dual of a Linear Transformation 143
The linear functional on U which has the effect [ff(y*)](a,) = a., is a{ip) =
XLi ? A If V = L"x *W«. then <%) = ^U *i(SLi ««A) = 25x (Si
bia ik )<f> k . Thus the representation of <r(y) is 5^. To follow absolutely the
notational conventions for representing a liner transformation as given in
Chapter II, (2.2), a should be represented by A T . However, because we
have chosen to represent y> by the row matrix B, and because a(y) is repre
sented by BA, we also use^ to represent a. We say that A represents a
with respect to B in V and A in U.
In most texts the convention to represent a by A T is chosen. The reason
we have chosen to represent a by A in this: in Chapter V we define a closely
related linear transformation a*, the adjoint of a. The adjoint is not repre
sented by A T ; it is represented by A* = A T , the conjugate complex of the
transpose. If we chose to represent a by A T , we would have a represented
by A, a by A T in both the real and complex case, and a* represented by A T
in the real case and A T in the complex case. Thus, the fact that the adjoint
is represented by A T in the real case does not, in itself, provide a compelling
reason for representing the dual by A T . There seems to be less confusion if
both a and a are represented by A, and a* is represented by A* (which
reduces to A T in the real case). In a number of other respects our choice
results in simplified notation.
If I g U, then y(<r(£)) = <%)(£), by definition of 6{xp). If f is represented
by JT, then v ( ff (D) = *UAT) = (5^ = d(y>)(f). Thus the representation
convention we are using allows us to interpret taking the dual of a linear
transformation as equivalent to the associative law. The interpretation could
be made to look better if we considered ffasa left operator on U and a right
operator on V. In other words, write cr(f) as ct and d(y>) as ypa. Then
V(fff) = (w)£ would correspond to passing to the dual.
Theorem 5.3. K(a) L = Im(o).
proof. If> e *(<7) c £ then for all a e U, v>(cr)a)) = a(v)(a) = 0. Thus
y e Imfa)^ If y, g ImCc) 1 , then for all aeU, <7(y)(a) = y(<*(a)) =
Thus y e /sT(o) and ^(5) = Im(a) 1 . D
Corollary 5.4. A necessary and sufficient condition for the solvability of
the linear problem <r(£) = p is that ft e ^(ct) 1 . n
The ideas of this section provide a simple way of proving a very useful
theorem concerning the solvability of systems of linear equations. The
theorem we prove, worded in terms of linear functional and duals, may not
at first appear to have much to do with with linear equations. But, when
worded in terms of matrices, it is identical to Theorem 7.2 of Chapter II.
144 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
Theorem 5.5. Let a be a linear transformation of U into V and let /? be any
vector in V. Either there is a £ 6 U such that
(1) <r(f) = 0,
or there is a <f> e V such that
(2) a(<t>) = 0and<f>p = l.
proof. Condition (1) means that (3 e Im(r) and condition (2) means that
/S £ K(6) L . Thus the assertion of the Theorem follows directly from Theorem
5.3. □
Theorem 5.5 is also equivalent to Theorem 7.2 of Chapter 2.
In matrix notation Theorem 5.5 reads: Let A be an m x n matrix and
B an m x 1 matrix. Either there is an n x 1 matrix X such that
(1) AX= B,
or there is a 1 X m matrix C such that
(2) CA = and CB = 1.
Theorem 5.6. a and a have the same rank.
proof. By Theorems 5.3 and 4.1, v{a) = n — p(a) = v(a). o
Theorem 5.7. Let W be a subspace of V invariant under a. Then W 1  is a
subspace of V invariant under a.
proof. Let </> 6 W L . For any a £ W we have a<f>(on.) = <f>o(aC) = 0, since
or(a) g W. Thus 6<j> e W ± . a
Theorem 5.8. The dual of a scalar transformation is also a scalar trans
formation generated by the same scalar.
proof. If ofa) = aa for all a e V, then for each cf> e V, (<7</>)(a) =0<r(a) =
cfyacx. = acfxx. □
Theorem 5.9. If A is an eigenvalue for a, then X is also an eigenvalue for a.
proof. If X is an eigenvalue for a, then a — X is singular. The dual of
a — X is a — X and it must also be singular by Theorem 5.6. Thus X is an
eigenvalue of a. □
Theorem 5.10. Let V have a basis consisting of eigenvectors of a. Then
V has a basis consisting of eigenvectors of a.
proof. Let {a 1} a 2 , . . . , a n } be a basis of V, and assume that o^ is an
eigenvector of a with eigenvalue X t . Let {</> l5 <f> 2 , . . . , <j> n ) be the corresponding
dual basis. For all a,, afai&j) = </>jff(a,) = ^iX^ctj = X i j> i c(. } = Xjd i:j = X^d^.
Thus a<j>i = X^i and fa is an eigenvector of a with eigenvalue X t . □
6  Duality of Linear Transformations 145
EXERCISES
1. Show that <tt = Tff.
2. Let crbea linear transformation of R 2 into R s represented by
R 1"
A =
2 4
2 2
Find a basis for (<r(R 2 )H. Find a linear functional that does not annihilate (12 1)
Show that (1,2, 1) £ <r(R 2 ).
3. The following system of linear equations has no solution. Find the linear
functional whose existence is asserted in Theorem 5.5.
$X^ f Xn ^ 2
x x + 2x 2 = 1
—x x + 3* 2 = 1.
*6 I Duality of Linear Transformations
In Section 5 we have defined the dual of a linear transformation. What is
the dual of the dual ? In considering this question we restrict our attention
to finite dimensional vector spaces. In this case, the mapping / of V into V,
defined in Section 2, is an isomorphism. Since a, the dual of a, is a mapping
of V into itself, the isomorphism J allows us to define a corresponding linear
transformation on V. For convenience, we also denote this linear transforma
tion by a. Thus,
d(cc) = /*[§(/(«))]. (6.1)
where the a on the left is the mapping of V into itself defined by the ex
pression on the right.
Theorem 6.1. The relation between a and a is symmetric; that is, a is
the dual of a.
proof. By definition,
<r(/(<x))(<£) = J(a)6(4) = dtfXa.) = <f>a(ai) = /(<r(a))(0).
Thus §(y(a)) = /(a(a)). By (6.1) this means a(a) = /"^(./(a))] =
J WffCa))] = <r(a). Hence, <r is the dual of a. D
The reciprocal nature of duality allows us to establish dual forms of theorems
without a new proof. For example, the dual form of Theorem 5.3 asserts
that K(oy = Im(ff). We exploit this principle systematically in this section.
Theorem 6.2. The dual of a monomorphism is an epimorphism. The
dual of an epimorphism is a monomorphism.
146 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
proof. By Theorem 5.3, Im(or) = K(a) L . If a is an epimorphism,
Im(<r) = V so that K(a) = V 1 = {0}. Dually, lm(a) = K(o) L . If a is a
monomorphism, K(a) = {0} and Im(a) = U. D
alternate proof. By Theorem 1.15 and 1.16 of Chapter II, a is an
epimorphism if and only if to = implies t = 0. Thus or = implies
f = if and only if a is an epimorphism. Thus a is an epimorphism if and
only if a is a monomorphism. Dually, r is a monomorphism if and only if
f is an epimorphism. □
Actually, a much more precise form of this theorem can be established.
If W is a subspace of V, the mapping i of W into V that maps <xeW onto
a e V is called the injection of W into V.
Theorem 6.3. Let W be a subspace of V and let i be the injection mapping
of W into V. Let R be the restriction map of V onto W. Then i and R are dual
mappings.
proof. Let (f>EV. For any a e W, i?(<£)(a) = <£i(a) = l(<f>)(a). Thus
R((f)) = \((/)) for each <j>. Hence, R = 1. D
Theorem 6.4. If it is a projection of U onto S along T, the dual rr is a
projection ofU onto T 1 along S L .
proof. A projection is characterized by the property tt 2 = tt. By
Theorem 5.7, ft 2 = tt* = ft so that tt is also a projection. By Theorem 5.3,
K(tt) = lminy = S ± and Im(tf) = K(tt) l = T 1 . □
A careful comparison of Theorems 6.2 and 6.4 should reveal the perils of
being careless about the domain and codomain of a linear transformation.
A projection tt of U onto the proper subspace S is not an epimorphism because
the codomain of tt is U, not S. Since if is a projection with the same rank as
77, 7r cannot be a monomorphism, which it would be if 77 were an epimorphism.
Theorem 6.5. Let be a linear transformation of U into V and let r be a
linear transformation of V into W. Let a and r be the corresponding dual
transformations. Iflm(a) = K(t), then Im(f) = K(a).
proof. Since Im(cr) <= K(t), rcr(a) = for all a e U; that is, to = 0.
Since or = to = 0, Im(f) c: K{6). Now dim Im(f) = dim Im(r) since r and
t have the same rank. Thus dim Im(r) = dim V — dim K{r) = dim V —
dim Im(cr) = dim V  dim lm(6) = dim K{6). Thus K(6) = Im(f). □
Definition. Experience has shown that the condition Im((r) = K{r) is very
useful because it is preserved under a variety of conditions, such as the
taking of duals in Theorem 6.5. Accordingly, this property is given a special
name. We say the sequence of mappings
U^>VL>W (6.1)
7  Direct Sums 147
is exact at V if Im(cr) = K{t). A sequence of mappings of any length is said
to be exact if it is exact at every place where the above condition can apply.
In these terms, Theorem 6.5 says that if the sequence (6.1) is exact at V, the
sequence
A. T A. # A.
U+~V^W (6.2)
is exact at V. We say that (6.1) and (6.2) are dual sequences of mappings.
Consider the linear transformation a of U into V. Associated with a is the
following sequence of mappings
_► K{a) U U !► V %► V/Im(a) — ► 0, (6.3)
where i is the injection mapping of K(a) into U, and »y is the natural homo
morphism of V onto V/Im(cr). The two mappings at the ends are the only
ones they could be, zero mappings. It is easily seen that this sequence is
exact.
Associated with a is the exact sequence
<— /Im(o) <— ^ V + K{a) <— 0. (6.4)
By Theorem 6.3 the restriction map R is the dual of i, and by Theorem 4.5
R an r\ differ by a natural isomorphism. With the understanding that
U/Im(a) is isomoprhic to K(a), and V/Im(<7) is isomorphic to K(a), the
sequences (6.3) and (6.4) are dual to each other.
*7 I Direct Sums
Definition. If A and B are any two sets, the set of pairs, (a, b), where a e A
and b e 6, is called the product set of A and 6, and is denoted by A x 6. If
i A i \i = 1 > 2, . . . , r]} is a finite indexed collection of sets, the product set of
the {A J is the set of all ntuples, (a lt a 2 , . . . , a n ), where a t e A f . This product
set is denoted by Xf =1 A t . If the index set is not ordered, the description of
the product set is a little more complicated. To see the appropriate generali
zation, notice that an «tuple in X» =1 A t , in effect, selects one element from
each of the A { . Generally, if {A tt \ f ie M} is an indexed collection of sets, an
element of the product set X„ eM A^ selects for each index fi an element of
A^. Thus, an element of X^ A^ is a function defined on M which associates
with each /<eA1an element a^ e A^.
Let {V { \i = 1, 2, ...,«} be a collection of vector spaces, all defined over
the same field of scalars F. With appropriate definitions of addition and
148 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
scalar multiplication it is possible to make a vector space over F out of the
product set X? =1 V^. We define addition and scalar multiplication as follows :
(a x , . . . , a n ) + (ft, . . . , ft) = (a x + ft, . . . , a M + ft) (7.1)
a(a l5 . . . , a„) = (aa l5 . . . , aa„). (7.2)
It is not difficult to show that the axioms of a vector space over F are satisfied,
and we leave this to the reader.
Definition. The vector space constructed from the product set Xj^y, by the
definitions given above is called the external direct sum of the V t and is
denoted by V 1 ® V 2 • • • ® V n = ©? =1 V*.
If D = ©f =1 Vj is the external direct sum of the V if the V t are not subspaces
of D (for n > 1). The elements of D are ntuples of vectors while the elements
of any V t are vectors. For the direct sum defined in Chapter I, Section 4, the
summand spaces were subspaces of the direct sum. If it is necessary to
distinguish between these two direct sums, the direct sum defined in Chapter I
will be called the internal direct sum.
Even though the V { are not subspaces of D it is possible to map the V {
monomorphically into D in such a way that D is an internal direct sum of these
images. Associate with a fc e V k the element (0, . . . , 0, a*, 0, . . . , 0) e D,
in which a fc appears in the kth position. Let us denote this mapping by i k .
i k is a monomorphism of V k into D, and it is called an injection. It provides
an embedding of V k in D. If V' k = lm(i k ) it is easily seen that D is an internal
direct sum of the V k .
It should be emphasized that the embedding of V k in D provided by the
injection map i k is entirely arbitrary even though it looks quite natural. There
are actually infinitely many ways to embed V k in D. For example, let a be any
linear transformation of V k into V x (we assume k ^ 1). Then define a new
mapping i' k of V k into D in which a fc e V k is mapped onto (<r(a fc ), 0, . . . , 0,
a w , 0, . . . , 0) eD. It is easily seen that i k is also a monomorphism of V k
into D.
Theorem 7.1. If dim U = m and dim V = n, then dim U ® V = m + n.
proof. Let A = {a l5 . . . , a OT } be a basis of U and let B = {ft, . . . , ft}
be a basis of V. Then consider the set {(a l5 0), . . . , (a m , 0), (0, ft),
(0, ft)} = (A, 8) in U ® V. If a = ^Zi *<«< and £ = 2?=i My. then
m n
(a, ft = 1^,0)+ 2 b,(0, ft)
and hence (A, B) spans U ® V. If we have a relation of the form
to n
2a i (a i ,0)+2^(0,ft) = 0,
i=i j=i
7  Direct Sums 149
then
1=1 3=1 /
and hence 2£i *<<** = ° and 2?=i V> = ° since A and 6 are linearl y in
dependent, all at = and all 6, : = 0. Thus (A, B) is a basis of U V and
1/ © V is of dimension m + n. D
It is easily seen that the external direct sum ©* =1 V it where dim V, = /w f , is
of dimension ]£> =1 m^
We have already noted that we can consider the field F to be a 1 dimen
sional vector space over itself. With this starting point we can construct
the external direct sum F ® F, which is easily seen to be equivalent to the
2dimensional coordinate space F 2 . Similarly, we can extend the external
direct sum to include more summands, and consider F n to be equivalent to
F ® • • • ® F, where this direct sum includes n summands.
We can define a mapping TT k of D onto V k by the rule ir k (oL lt . . . , <x n ) = a fc .
7T k is called a projection of D onto the kth component. Actually, rr k is not a
projection in the sense of the definition given in Section II 1, because here
the domain and codomain of n k are different and 7r fc 2 is not defined. However,
(i k 7T k y = t k 7T k i k TT k = i k \7T k = i k n k so that t fc 7T fc is a. projection. Let W k denote
kernel of 7T k . It is easily seen that
W k = V 1 © • • • © Vj, © {0} © V k+l © • • • © V n . (7.3)
The injections and projections defined are related in simple but important
ways. It is readily established that
"A = U k , 7  4 )
n.^ = for ijLk, (7.5)
tjTT! + • • ' + L n TT n = 1 D . (7.6)
The mappings i k 7r t for i 9^ k are not defined since the domain of i k does not
include the codomain of 77^.
Conversely, the relation (7.4), (7.5), and (7.6), are sufficient to define the
direct sum. Starting with the V k , the monomorphisms i k embed the V k in D.
Let V k = Im(« fc ). Let D' = V[ + • • • + V;. Conditions (7.4) and (7.5) imply
that D' is a direct sum of the V k . For if = a( + • • • + <, with c/.' k e V^,
there exist a fc g V fc such that i k (<x k ) = cn k . Then 7r fc (0) = 7r fc (a() + • • • +
^*«) = 7Vi(*i) + ' ' • + T^CaJ = a fc = 0. Thus < = and the sum is
direct. Condition (7.6) implies that D' = D.
A A.
Theorem 7.2. The dual space of U © V « naturally isomorphic to U © V.
proof. First of all, if dim U = m and dim V = n, then dim (7 © V =
m + n and dim U ®V = m + n. Since U © V and U © V have the same
150 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
dimension, there exists an isomorphism between them. The real content
of this theorem, however, is that this isomorphism can be specified in a
natural way independent of any coordinate system.
For ((f>, y>) e © V and (a, P)eU®V, define
(cf>, vO(a, 0) = <£a + y>0. (7.7)
It is easy to verify that this mapping of (a, /?) s U ® V onto </>a + ip{3 e F is
linear and, therefore, corresponds to a linear functional, an element of
U ® V. It is also easy to verify that the mapping of U ® V into U ® V that this
defines is a linear mapping. Finally, if (<j>, ip) corresponds to the zero linear
functional, then (c/>, tp)(<x, 0) = (fxx. = for all oleU. This implies that <f> = 0.
In a similar way we can conclude that y> = 0. This shows that the mapping of
A A ^^^\
U © V into l) © Vhas kernel {(0, 0)}. Thus the mapping is an isomorphism. □
Corollary 7.3. The dual space to V x ® • • • © V n is naturally isomorphic
to V x © • • • © V n . □
The direct sum of an infinite number of spaces is somewhat more com
plicated. In this case an element of the product set P = X MeM V is a function
on the index set M. For a e X^V^, let a,, = <x(/u) denote the value of this
function in V„. Then we can define a + (3 and a<x (for a e F) by the rules
(a + P)(p) = a„ + ^, (7.8)
(fla)Cu) = aa^. (7.9)
It is easily seen that these definitions convert the product set into a vector
space. As before, we can define injective mappings i^ of V M into P. However,
P is not the direct sum of these image spaces because, in algebra, we permit
sums of only finitely many summands.
Let D be the subset of P consisting of those functions that vanish on all but a
finite number of elements of M. With the operations of vector addition and
scalar multiplication defined in P, D is a subspace. Both D and P are useful
concepts. To distinguish them we call D the external direct sum and P the
direct product. These terms are not universal and the reader of any mathe
matical literature should be careful about the intended meaning of these or
related terms. To indicate the summands in P and D, we will denote P by
X^V, and D by 0^ V„.
In a certain sense, the external direct sum and the direct product are dual
concepts. Let i denote the injection of V into P and let tt denote the pro
jection of P onto V . It is easily seen that we have
and
7  Direct Sums 151
These mappings also have meaning in reference to D. Though we use the
same notation, n^ requires a restriction of the domain and i^ requires a
restriction of the codomain. For D the analog of (7.6) is correct,
Even though the left side of (7.6)' involves an infinite number of terms,
when applied to an element a g D,
( 2 ^)(«) = 2 («„*■„)(«) = 2 W = a ( 7  10 )
jieAl /ie/V1 Me/V1
involves only a finite number of terms. An analog of (7.6) for the direct
product is not available.
Consider the diagram of mappings
V^D^V V , (7.11)
and consider the dual diagram
$ u JldJ1V v . (7.12)
For v 5^ fi, TTyip = 0. Thus t M fi, = ir^ = 0. For v = p, t^ = tt^ =
1 = 1. By Theorem 6.2, t^ is an epimorphism and tt^ is a monomor
phism. Thus fr^ is an injection of V^ into D, and t^ is a projection of D
onto V„.
Theorem 7.4. If D is the external direct sum of the indexed collection
{V fi \ /j, e M}, D is isomorphic to the direct product of the indexed collection
proof. Let (f>G D. For each fie M, fa^ is a linear functional defined on
V^; that is, ^ corresponds to an element in V M . In this way we define a
function on M which has at/teM the value </>t„ e 9^. By definition, this is
an element in X MsM V M . It is easy to check that this mapping of D into the
direct product X^ gA1 V^ is linear.
If (f> ?£ 0, there is an a e D such that <f><x. ^ 0. Since <f><x = (f> [Q^e/w V^*)] —
2i"6/vi <£ V7*( a ) ^ ^» tnere * s a A* G M such that fapTT^cn.) ^ 0. Since
7^ (a) g V^, <£^ # 0. Thus, the kernel of the mapping of D into X^ eM V,, is
zero.
a.
Finally, we show that this mapping is an epimorphism. Let y> e X^ eA1 V^.
Let ^ = y(/*) g V^ be the value of y> at /*. For a g D, define 0a =
2 i «6AiV , / < ( 7r A* a ) This sum is defined since tt^ol = for all but finitely many ft.
152 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
For a„ e V v ,
<KOv) = $> v a v )
= V»(«v) (7.13)
This shows that y> is the image of <f>. Hence, D and X fieM V^ are ismorphic. n
While Theorem 7.4 shows that the direct product 6 is the dual of the exter
nal direct sum D, the external direct sum is generally not the dual of the direct
product. This conclusion follows from a fact (not proven in this book) that
infinite dimensional vector spaces are not reflexive. However, there is more
symmetry in this relationship than this negative assertion seems to indicate.
This is brought out in the next two theorems.
Theorem 7.5. Let {V„  fi e M} be an indexed collection of vector spaces
over F and let {a^ \ /u, e M} be an indexed collection of linear transformations,
where <r„ has domain V^ and codbmain Ufor all pi. Then there is a unique linear
transformation a o/©^^ into U such that a^ = ai^ for each ju.
proof. Define
a = 2 <W (7.14)
HeM
For each a e ® /ieM v », *(<*) = 2^m V7«( a ) is well defined since only a finite
number of terms on the right are nonzero. Then, for a„ e V v ,
= 2 o^/vXav)
fieM
= <W (7.15)
Thus ai v = a v .
If a' is another linear transformation of ® MeM V„ into U such that <r„ = a'l^,
then
a' = a'\
D
lieM
= 2 a'l 77,,
= 2 <W
/16M
= O".
Thus, the a with the desired property is unique. □
7  Direct Sums 153
Theorem 7.6. Let {V„  y. e M} fee an indexed collection of vector spaces
over F and let {r^  fx e M} fee an i/Ktaw* collection of linear transformations
where t„ /ias ^omaw W and codomain V„ /or a// /*. 77k?h fAere is a linear
transformation rofW into X^ V„ swcA */zcrt t^ = ir^for each [x.
proof. Let a g W be given. Since r(a) is supposed to be in X^V^,
r(a) is a function on M which for ^ g M has a value in V„. Define
T(a)0«) = t». (716)
Then
^r(a) = T(a)(^) = t„(«), (7.17)
SO that 77^ T = T^. D
The distinction between the external direct sum and the direct product is
that the external direct sum is too small to replace the direct product in
Theorem 7.6. This replacement could be done only if the indexed collection
of linear transformations were restricted so that for each a e W only finitely
many mappings have nonzero values ^(a).
The properties of the external direct sum and the direct product established
in Theorems 7.5 and 7.6 are known as "universal factoring" properties. In
Theorem 7.5 we have shown that any collection of mappings of V„ into a space
U can be factored through D. In Theorem 7.6 we have shown that any
collection of mappings of W into the V„ can be factored through P. Theorems
7.7 and 7.8 show that D and P are the smallest spaces with these properties.
Theorem 7.7. Let W be a vector space over F with an indexed collection
of linear transformations {X M \ /x e M} where each X^ has domain V^ and co
domain W. Suppose that, for any indexed collection of linear transformations
{a  fx g M} with domain V^ and codomain U, there exists a linear transforma
tion XofW into U such that o^ = XX^. Then there exists a monomorphism of
D into W.
proof. By assumption, there exists a linear transformation X of W into
D such that ^ = XX,,. By Theorem 7.5 there is a unique linear transformation
a of D into W such that X^ = oi„. Then
= ^ XX^TT^
= Xo 2 l^TTp
= Xo. (7.18)
This means that a is a monomorphism and X is an epimorphism. □
154 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
Theorem 7.8. Let Y be a vector space over F with an indexed collection
of linear transformations {Q fl \[ie M} where each 6^ has domain Y and codomain
V M . Suppose that, for any indexed collection of linear transformations
{ T ix I f* e M} with domain W and codomain V^, there exists a linear transfor
mation OofW into Y such that r^ = 6^6. Then P is isomorphic to a subspace
o/Y.
proof. With P in place of W and ^ in place of t„, the assumptions
of the theorem say there is a linear transformation of P into Y such that
^n = OpO for each fi. By Theorem 7.6 there is a linear transformation r of V
into P such that 9^ = tt^t for each [x. Then
^ = OpO = wyrfl.
Recall that a e P is a function defined on M that has at ju e M a value a
in V M . Thus a is uniquely defined by its values. For /u e M
ff„(T0(a)) = ^(a) = a„.
Thus r0(a) = a and t6 = 1 P . This means that 6 is a monomorphism and r
is an epimorphism and P is isomorphic to lm(0). □
Theorem 7.9. Suppose a space D' is given with an indexed collection of
monomorphisms {^  /u, e M} of V^ into D' and an indexed collection of epi
morphisms {tt^\ jug M} ofD' onto V M such that
*'/». = 1/ m
77
' v ip = for v ?£ /u,
Z, l li 7T fi = Id'
fieM
Then D and D' are isomorphic.
This theorem says, in effect, that conditions (7.4), (7.5), and (7.6)'
characterize the external direct sum.
proof. For a e D' let a„ = w^(a). We wish to show first that for a given
a e D' only finitely many a„ are nonzero. By (7.6)' a = l D (a) =
X«e/M *^(«) = X«eM ^<V Thus, only finitely many of the t^ are nonzero.
Since ^ is a monomorphism, only finitely many of the a„ are nonzero.
Now suppose that {a^  ju e M} is an indexed collection of linear transforma
tions with domain V„ and codomain U. Define X = ^ eM o^. For a e D',
^( a ) = X«e/n ce fi 7T ' f i(< x ) = Xue/vi a ^n * s defined in U since only finitely many a^
are nonzero. Also, Xi'^ = Q> eM <VvK = V TmiS D ' satisfies the condi
tions of W in Theorem 7.7.
Repeating the steps of the proof of Theorem 7.7, we have a monomorphism
a of D into D' and an epimorphism X of D' onto D such that 1 D = Xa. But
7  Direct Sums
we also have
155
Id' = 2 l 'n n ii
tieM
lieM
= <tA.
Since cr is both a monomorphism and an epimorphism, D and D' are iso
morphic. □
The direct product cannot be characterized quite so neatly. Although
the direct product has a collection of mappings satisfying (7.4) and (7.5),
(7.6)' is not satisfied for this collection if M is an infinite set. The universal
factoring property established for direct products in Theorem 7.6 is inde
pendent of (7.4) and (7.5), since direct sums satisfy (7.4) and (7.5) but not
the universal factoring property of Theorem 7.6. We can combine these three
conditions and state the following theorem.
Theorem 7.10. Let P' be a vector space over F with an indexed collection of
monomorphisms {i^\/*e M} of V^ into P' and an indexed collection of epi
morphisms «  jm e M} ofP' onto V M such that
77;^ = for r^jw
and such that if{ Pli \ fx e M} is any indexed collection of linear transformations
with domain W and codomain V„, there is a linear transformation pofW into
P' such that Pfl = ir'rffor each fi. IfP' is minimal with respect to these three
properties, then P and ?' are isomorphic.
When we say that P' is minimal with respect to these three properties we
mean: Let P" be a subspace of P' and let wj be the restriction of < to P".
If there exists an indexed collection of monomorphisms {t£  n e M} with
domain V„ and codomain P" such that (7.4), (7.5) and the universal factoring
properties are satisfied with tj in place of ^ and tjJ in place of *£, then P" = ?'.
proof. By Theorem 7.8, P is isomorphic to a subspace of ?'. Let be the
isomorphism and let P" = lm(0). With appropriate changes in notation
(P' in place of Y and tt; in place of 6 J, the proof of Theorem 7.8 yields
the relations
^u = 7r u T »
156 . Linear Functionals, Bilinear Forms, Quadratic Forms  IV
where t is an epimorphism of ?' onto P. Thus, if ttJ is the restriction of n'^
to P", we have
This shows that 77^' is an epimorphism.
Now let £ = 0^.
11.1t irt\. a
and
ff vC = t^V = ""V/* = for v ^ [jl.
Since P has the universal factoring property, let r be a linear transformation
of W into P such that />,, = ir^r for each ^. Then
P, = *V = <^r = Try
for each ^, where r" = 0r. This shows that P" has universal factoring
property of Theorem 7.6. Since we have assumed ?' is minimal, we have
P" = P' so that P and P' are isomorphic. D
8 I Bilinear Forms
Definition. Let U and V be two vector spaces with the same field of scalars
F. Let/ be a mapping of pairs of vectors, one from U and one from V, into
the field of scalars such that /(a, p), where <x.eil and p g V, is a linear
function of a and ^ separately. Thus,
/(a iai + * 2 a 2 , ^ + 6 2 &) = ai /( ai , Ajft + 6 2 &) + a 2 /(a 2 , ^ + 6 2 /? 2 )
= aA/(a l5 &) + aA/(a x , p 2 )
+ «2^ 1 /(a 2 , A) + a 2 b 2 f(a 2 , p 2 ). (8.1)
Such a mapping is called a bilinear form. In most cases we shall have U = V.
(1) Take U = V = R n and F = R. Let A = {a l5 . . . , aj be a basis in
R n . For I = JjLx *i«< and /? = 2» =1 */*** we may defme/(, rj) = 2» =1 *#,.
This is a bilinear form and it is known as the inner, or dot, product.
(2) We can take F = R and U = V = space of continuous realvalued
functions on the interval [0, 1]. We may then define/(a, P) = JJ *(x)P(x) dx.
This is an infinite dimensional form of an inner product. It is a bilinear form.
As usual, we proceed to define the matrices representing bilinear forms
with respect to bases in U and V and to see how these matrices are transformed
when the bases are changed.
Let A = {on, . . . , a w } be a basis in U and let 8 = {&, . . . , p n } be a basis
in V. Then, for any <x e U, e V, we have a = J~ 1 x^ and p = 2? =1 y,P f
8  Bilinear Forms
where x if y i e F. Then
157
m / n \
= I2¥^.^ (82)
1=1 3=1
Thus we see that the value of the bilinear form is known and determined
for any a £ U, ft g V, as soon as we specify the mn values /(oc i5 /?,). Con
versely, values can be assigned to/(oc l5 ft } ) in an arbitrary way and /(a, /?)
can be defined uniquely for all a e U, {3 eV, because A and B are bases in
U and V, respectively.
We denote/(a i5 ft ) by & i3 and define B = [b^] to be the matrix represent
ing the bilinear form with respect to the bases A and 8. We can use the
ratuple X = {x y , . . . , x m ) to represent a and the ntuple Y = (y u . . . , y n )
to represent ft. Then
/(a, ft) = ^ 2 *A;2/;
i=i j=i
= [*!••• * J5
= * t by\
yi
LyJ
(8.3)
(Remember, our convention is to use an mtuple X = (x u . . . , x m ) to
represent an m x 1 matrix. Thus X and Y are onecolumn matrices.)
Suppose, now, that A' — {<*.[, . . . , cn' m }is a new basis of U with matrix
of transition P, and that B' = {ft[, . . . , ft^} is a new basis of V with matrix
of transition Q. The matrix B' = [b' i} ] representing / with respect to these
new bases is determined as follows :
«m n \
Z Pri*r 2 «.y&)
r=l s=l /
= J,Pr
r=l
2 Qsif(*r. Ps)
= 22 PriKsQsi
r=l s=l
(8.4)
158 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
Thus,
B' = P T BQ. (8.5)
From now on we assume that U = V. Then when we change from one
basis to another, there is but one matrix of transition and P = Q in the
discussion above. Hence a change of basis leads to a new representation
of/ in the form
B' = P T BP. (8.6)
Definition. The matrices B and P T BP, where P is nonsingular, are said to
be congruent.
Congruence is another equivalence relation among matrices. Notice
that the particular kind of equivalence relation that is appropriate and
meaningful depends on the underlying concept which the matrices are
used to represent. Still other equivalence relations appear later. This
occurs, for example, when we place restrictions on the types of bases we
allow.
Definition. If /(a, 0) = /(/?, a) for all a, /? e V, we say that the bilinear form
/is symmetric. Notice that for this definition to have meaning it is necessary
that the bilinear form be defined on pairs of vectors from the same vector
space, not from different vector spaces. If/ (a, a) = for all a e V, we say
that the bilinear form /is skewsymmetric.
Theorem 8.1. A bilinear form f is symmetric if and only if any matrix B
representing f has the property B T = B.
proof. The matrix B = [b i} ] is determined by /(a i5 a,). But b H =
f(cL it oO = /(a„ a,) = b i} so that B T = B.
If B T = B, we say the matrix B is symmetric. We shall soon see that
symmetric bilinear forms and symmetric matrices are particularly important.
If B T = B, then /(a„ a y ) = b ti = b H =/(a„ a,). Thus /(a, p) =
/(I?=i «,«!, Iu b >*>) = 2Li 2"i W(«<» «*) = 2?=i 2"i W(«*. a *) =
/(/?, a). It then follows that any other matrix representing/will be symmetric ;
that is, if B is symmetric, then P T BP is also symmetric. □
Theorem 8.2. If a bilinear form f is skewsymmetric, then any matrix B
representing f has the property B T = —B.
proof. For any a, e V, = /(a + & a + 0) = /(«., a) + /(a, 0) +
/Off, a) +/(/8, £) =/(a, /8) +/(/9, a). From this it follows that /(a, 0) =
/(£, a) and hence B T = —B.U
Theorem 8.3. If 1 + 1 5* ana? f/ie matrix B representing f has the
property B T = — B, then f is skewsymmetric.
8  Bilinear Forms 1*9
proof. Suppose that B T = B, or /(a, P) = f(B, a) for all a, e V.
Then /(a, a) = /(a, a), from which we have /(a, a) +/(<x, a) =
(1 + l)/(a, a) = 0. Thus, if 1 + 1 ^ 0, we can conclude that /(a, a) =
so that /is skewsymmetric. □
If B T = —B, we say the matrix B is skewsymmetric. The importance
of symmetric and skewsymmetric bilinear forms is implicit in
Theorem 8.4. If 1 + 1^0, every bilinear form can be represented
uniquely as a sum of a symmetric bilinear form and a skewsymmetric bilinear
form.
proof. Let /be the given bilinear form. Define / s ( a, P) = [/(a, P) +
/{0,a)]and/„Ox,0) = il/(a, j8) /(jff,a)]. (The assumption that 1 + 1 *
is required to assure that the coefficient "£" has meaning.) It is clear that
/ s (a, fl =/,(j0, a) and/ ss (a, a) = so that/ is symmetric and/„ is skew
symmetric.
We must yet show that this representation is unique. Thus, suppose that
/( a , p) =/ x (a, P) +/ 2 (a, P) where /. is symmetric and/ 2 is skewsymmetric.
Then /(a, P)+f{$, a) =/ 1 (a, 0) . + /,(«, P)+fi(fi, «)+/■(& «) =
2/x(a, 0). Hence /i(a, 0) = £/(a, 0) + /(& a)]. If follows immediately
that/ 2 (a, 0) = £[/(a, 0) /(£, a)]. D
We shall, for the rest of this book, assume that 1 + 1^0 even where
such an assumption is not explicitly mentioned.
EXERCISES
1. Let a = (» x , x 2 ) g R 2 and let j8 = (y lt y 2 , y 3 ) e R 3 . Then consider the bilinear
form
/(a, 0) = a;^ + 2x^2  a^!  x 2 y 2 + 6x x t/ 3 .
Determine the 2 x 3 matrix representing this bilinear form.
2. Express the matrix
1 2 3"
4 5 6
7 8 9
as the sum of a symmetric matrix and a skewsymmetric matrix.
3. Show that if B is symmetric, then P T BP is symmetric for each P, singular or
nonsingular. Show that if B is skewsymmetric, then P^BP is skewsymmetric
for each P.
4. Show that if A is any m x n matrix, then A T A and AA T are symmetric.
5. Show that a skewsymmetric matrix of odd order must be singular.
160 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
6. Let / be a bilinear form defined on U and V. Show that, for each a £ I/,
/(a, /?) defines a linear functional <f> a on V; that is,
*«(#=/(«, 0).
With this fixed / show that the mapping of a £ U onto <f> a £ V is a linear trans
formation of U into V.
7. (Continuation) Let the linear transformation of U into V defined in Exercise
6 be denoted by a f . Show that there is an a £ U, a ^ 0, such that /(a, /J) = for
all jS if and only if the nullity of a f is positive.
8. (Continuation) Show that for each /? £ V, /Y<x, p) defines a linear function \p*
on U. The mapping of p £ V onto y^ e L/ is a linear transformation ^ of V into L/.
9. (Continuation) Show that a f and r f have the same rank.
10. (Continuation) Show that, if U and V are of different dimensions, there
must be either an a £ U, a ^ 0, such that /(a, /S) = for all p £ V or a /? e V,
/? t^ 0, such that /(a, /5) = for all a £ U. Show that the same conclusion follows
if the matrix representing /is square but singular.
11. Let U be the set of all a £ U such that /(a, 0) = for all peV. Similarly,
let V be the set of all £ V such that /(a, 0) = for all a £ U. Show that U is a
subspace of U and that V is a subspace of V.
12. (Continuation) Show that m — dim U = n — dim V .
13. Show that if / is a skewsymmetric bilinear form, then /(a, /?) = — /(/?, a)
for all a, peV.
14. Show by an example that, if A and B are symmetric, it is not necessarily true
that AB is symmetric. What can be concluded if A and B are symmetric and
AB = BAt
15. Under what conditions on B does it follow that X T BX = for all XI
16. Show the following: If A is skewsymmetric, then A 2 is symmetric. If A is
skewsymmetric and B is symmetric, then AB — BA is symmetric. If A is skew
symmetric and B is symmetric, then AB is skewsymmetric if and only if AB = BA.
9 I Quadratic Forms
Definition. A quadratic form is a function q on a vector space defined by
setting #(<x) =/(a, a), where/ is a bilinear form on that vector space.
If/ is represented as a sum of a symmetric and a skewsymmetric bilinear
form, /(a, 0) =/ s (a, 0) +/ ss (a, 0) where/, is symmetric and /„ is skew
symmetric, then^r(a) =/ s (a, a) +/ ss (a, a) =/ s (a, a). Thus # is completely
determined by the symmetric part of/ alone. In addition, two different
bilinear forms with the same symmetric part must generate the same quadratic
form.
We see, therefore, that if a quadratic form is given we should not expect
9  Quadratic Forms I" 1
to be able to specify the bilinear form from which it is obtained. At best
we can expect to specify the symmetric part of the underlying bilinear form.
This symmetric part is itself a bilinear form from which q can be obtained.
Each other possible underlying bilinear form will differ from this symmetric
bilinear form by a skewsymmetric term.
What is the symmetric part of the underlying bilinear from expressed
in terms of the given quadratic form ? We can obtain a hint of what it should
be by regarding the simple quadratic function x 2 as obtained from the bilinear
function xy. Now (x + yf = x 2 + xy + yx + y 2 . Thus if xy = yx (sym
metry), we can express xy as a sum of squares, xy = \[(x + y*) — x 2 — y 2 ].
In general, we see that the symmetric part of the underlying bilinear form
can be recovered from the quadratic form by means of the formula
= *[/"(<* + ft a + 0) /(a, <x)/(ft ft]
= *[/"(«, «) +/(«, +/(/?, a) + /(ft /(«, «) /(ft ft]
= il/(a,j8)+/(ft<x)]
= />,ft ( 9  1}
/, is the symmetric part of/. Thus it is readily seen that
Theorem 9.1. Every symmetric bilinear form f determines a unique
quadratic form by the rule q(a) =/(«, a), and if 1 + 1 ^ 0, every quadratic
form determines a unique symmetric bilinear form / s (a, ft = \[q{* + ft —
^( a ) — q(fi)]from which it is in turn determined by the given rule. There is a
onetoone correspondence between symmetric bilinear forms and quadratic
forms. □
The significance of Theorem 9.1 is that, to treat quadratic forms ade
quately, it is sufficient to consider symmetric bilinear forms. It is fortunate
that symmetric bilinear forms and symmetric matrices are very easy to
handle. Among many possible bilinear forms corresponding to a given
quadratic form a symmetric bilinear form can always be selected. Hence,
among many possible matrices that could be chosen to represent a given
quadratic form, a symmetric matrix can always be selected.
The unique symmetric bilinear form/ s obtainable from a given quadratic
form q is called the polar form of q.
It is desirable at this point to give a geometric interpretation of quadratic
forms and their corresponding polar forms. This application of quadratic
forms is by no means the most important, but it the source of much of the
terminology. In a Euclidean plane with Cartesian coordinate system, let
(x) = (x lt x t ) be the coordinates of a general point. Then
q(( x )) = x \ — ^ x i x * + 2x z 2
162 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
is a quadratic function of the coordinates and it is a particular quadratic
form. The set of all points (x) for which q((x)) = 1 is a conic section (in
this case a hyperbola).
Now, let (y) = (y x , y 2 ) be the coordinates of another point. Then
/«((*), 0)) = X \V\  2x lV2  2x 2Vl + 2 *22/2
is a function of both (x) and (y) and it is linear in the coordinates of each
point separately. It is a bilinear form, the polar form of q. For a fixed
(x), the set of all (y) for which f s ((x), (y)) = 1 is a straight line. This straight
line is called the polar of (x) and (x) is called the pole of the straight line.
The relations between poles and polars are quite interesting and are ex
plored in great depth in projective geometry. One of the simplest relations
is that if (x) is on the conic section defined by q((x)) = 1 , then the polar of
(x) is tangent to the conic at (x). This is often shown in courses in analytic
geometry and it is an elementary exercise in calculus.
We see that the matrix representing/ s ((a;), (y)), and therefore also q((x)), is
1 2"
EXERCISES
1. Find the symmetric matrix representing each of the following quadratic
forms :
(a) 2x 2 + 3xy + 6y 2
(b) %xy + 4y 2
(c) x 2 + 2xy + Axz + 2>y 2 + yz + lz 2
(d) Axy
(e) x 2 + Axy + Ay 2 + 2xz + z 2 + Ayz
(/) x 2 + Axy — 1y 2
(g) x 2 + 6xy — 2y 2 — 2yz + z 2 .
2. Write down the polar form for each of the quadratic forms of Exercise 1 .
3. Show that the polar form/ s of the quadratic form q can be recovered from the
quadratic form by the formula
/,(«, P) = i{?(« + P) ~ ?(«  P)}
10 I The Normal Form
Since the symmetry of the polar form f s is independent of any coordinate
system, the matrix representing f s with respect to any coordinate system
will be symmetric. The simplest of all symmetric matrices are those for
which the elements not on the main diagonal are all zeros, the diagonal
matrices. A great deal of the usefulness and importance of symmetric
10 I The Normal Form
163
bilinear forms lies in the fact that for each symmetric bilinear form, over
a field in which 1 + 1^0, there exists a coordinate system in which the
matrix representing the symmetric bilinear form is a diagonal matrix. Neither
the coordinate system nor the diagonal matrix is unique.
Theorem 10.1. For a given symmetric matrix B over a field F (in which
1 + 1^0), there is a nonsingular matrix P such that P T BP is a diagonal
matrix. In other words, if f is the underlying symmetric bilinear (polar)
form, there is a basis A' = «, ...,<} of V such thatf s (^, a)). = whenever
i 5*/
proof. The proof is by induction on n, the order of B. If n = 1, the
theorem is obviously true (every 1 x 1 matrix is diagonal). Suppose the
assertion of the theorem has already been established for a symmetric
bilinear form in a space of dimension n — 1. If B = 0, then it is already
diagonal. Thus we may as well assume that B ^ 0. Let f s and q be the
corresponding symmetric bilinear and quadratic forms. We have already
shown that
/.(«, P) = ifo(* + P) ~ ?(<*)  ?(#]• (1(U)
The significance of this equation at this point is that if q(<x.) = for all a, ^
then/ s (a, 0) = for all a and 0. Hence, there is anajeV such that 4(04) = ^J^
d 1 ^0. ' . /?■".'
With this 04 held fixed, the bilinear form/,(a;, a) defines^ linear functional, f s
<f>[ on V. This linear functional is not zero since <£X = d x 5* 0. Thus the
subspace W x annihilated by this linear functional is of dimension n  1.
Consider f restricted to VVJ This is a symmetric bilinear form on W x j {
and, by assumption, there is a basis {a;, . . . , <} of W x such that/ s (a^, aj) =
if i *j and 2 < 1, ; < n. However, / s «, <) = / s K, <) = because of s j,
symmetry and the fact that a ; e W 1 for / > 2. Thus/ S «, aj) = if/ 5* ; for f
1 < *, ;<«•□
Let P be the matrix of transition from the original basis A = {a 1? . . . , a„}
to the new basis A' = {aj, . . . , <}. Then P T 5i> = B' is of the form
{% Oi
B' =
Vi
rf 2
In this display of B' the first r elements of the main diagonal are nonzero
164
Linear Functional, Bilinear Forms, Quadratic Forms  IV
and all other elements of B' are zero, r is the rank of B' and B, and it is
also called the rank of the corresponding bilinear or quadratic form.
The di$ along the main diagonal are not uniquely determined. We can
introduce a third basis A" = {cd' x , . . . , ct." n } such that a" = a^a! where x t ^ 0.
Then the matrix of transition Q from the basis A' to the basis A" is a diagonal
matrix with x lt . . . , x n down the main diagonal. The matrix B" representing
the symmetric bilinear form with respect to the basis A" is
B" = Q T B'Q =
d x x x
W..X*.
.
Thus the elements in the main diagonal may be multiplied by arbitrary
nonzero squares from F.
"2 1"
1 2
we
get B" = P T B'P =
By taking B' = and P =
■3 01 Lo ij
. Thus, it is possible to change the elements in the main diagonal
by factors which are not squares. However, \B"\ = \B'\ • \P\ 2 so that it
is not possible to change just one element of the main diagonal by a non
square factor. The question of just what changes in the quadratic form can
be effected by P with rational elements is a question which opens the door
to the arithmetic theory of quadratic forms, a branch of number theory.
Little more can be said without knowledge of which numbers in the field
of scalars can be squares. In the field of complex numbers every number
is a square; that is, every complex number has at least one square root.
Therefore, for each d t ^ we can choose x i = —= so that djX? = 1.
y/di
In this case the nonzero numbers appearing in the main diagonal of B"
are all l's. Thus we have proved
Theorem 10.2. If F is the field of complex numbers, then every symmetric
matrix B is congruent to a diagonal matrix in which all the nonzero elements
are Vs. The number of Vs appearing in the main diagonal is equal to the
rank of B. □
The proof of Theorem 10.1 provides a thoroughly practical method for find
ing a nonsingular P such that P T BP is a diagonal matrix. The first problem
10 I The Normal Form 165
is to find an <x[ such that q(a' x ) ^ 0. The range of choices for such an «.[ is
generally so great that there is no difficulty in finding a suitable choice by
trial and error. For the same reason, any systematic method for finding
an 04 must be a matter of personal preference.
Among other possibilities, an efficient system for finding an a[ is the
following: First try < = a x . If qM = b u = 0, try < = a 2 . If ?(a 2 ) =
b 2% = 0,thenqr( ai + a 2 ) = qM + 2/ s ( ai , a 2 ) + ?(a 2 ) = 2/ s ( ai a 2 ) = 2b 12 so
that it is convenient to try aj = a x + a 2 . The point of making this sequence
of trials is that the outcome of each is determined by the value of a single
element of B. If all three of these fail, then we can pass our attention to
<x 3 , a x + a 3 , and oc 2 + a 3 with similar ease and proceed in this fashion.
Now, with the chosen *[, /,«, a) defines a linear functional <j>[ on V.
If 04 is represented by (p n , . . . , p nX ) and a by (z x , . . . , *„), then
n n n / n \
/ s (al, a) = 2 2 PaM* =22 PabnW (10.2)
This means that the linear functional <f>[ is represented by [p n • • • p n i]5.
The next step described in the proof is to determine the subspace V^
annihilated by <f>' r However, it is not necessary to find all of W x . It is
sufficient to find an <x 2 e W x such that q(a.' 2 ) 7* 0. With this a;, /,(a£, a)
defines a linear functional <£ 2 on V. If a 2 is represented by (p 12 , . . . , /7„ 2 ),
then <j>' 2 is represented by [p 12 • • • /? n2 ]5
The next subspace we need is the subspace W 2 of W l annihilated by <f> 2 .
Thus W 2 is the subspace annihilated by both cf>[ and 0;. We then select an
ol' 2 from W 2 and proceed as before.
Let us illustrate the entire procedure with an example. Consider
"0 1 2"
B =
1 1
2 10.
Since b n = b 22 = 0, we take a{ = a x + a 2 = (1, 1, 0). Then the linear
functional fa is represented by
[1 1 0]B = [1 1 3].
A possible choice for an 04 annihilated by this linear functional is (1 , —1,0).
The linear functional fa determined by (1, 1, 0) is represented by
[1 _i (J]f}=[1 1 1].
We should have checked to see that q{vQ 5^ 0, but it is easier to make that
check after determining the linear functional fa 2 since q(u.' 2 ) = fa 2 a.' 2 =
—2^0 and the arithmetic of evaluating the quadratic form includes all
the steps involved in determining fa 2 .
166
Linear Functionals, Bilinear Forms, Quadratic Forms  IV
We must now find an 0C3 annihilated by (f>[ and <f>' 2 . This amounts to solving
the system of homogeneous linear equations represented by
1 1 3"
1 1 1
A possible choice is a.' 3 = ( — 1, —2, 1). The corresponding linear func
tional (^3 is represented by
[1 2 l]fl=[0 4].
The desired matrix of transition is
1 1 r
P =
1 1
1
1
1
3]
1
1
1
4J
1
1
f
1
1
2
=
1_
2
0"
2
.0
4.
Since the linear functionals we have calculated along the way are the rows
of P T B, the calculation of P T BP is half completed. Thus,
P T BP =
It is possible to modify the diagonal form by multiplying the elements in
the main diagonal by squares from F. Thus, if F is the field of rational
numbers we can obtain the diagonal {2, —2, —1}. If F is the field of real
numbers we can get the diagonal {1,1,1}. If F is the field of com
plex numbers we can get the diagonal {1, 1, 1}.
Since the matrix of transition P is a product of elementary matrices the
diagonal from P T BP can also be obtained by a sequence of elementary
row and column operations, provided the sequence of column operations
is exactly the same as the sequence of row operations. This method is
commonly used to obtain the diagonal form under the congruence. If an
element b u in the main diagonal is nonzero, it can be used to reduce all other
elements in row / and column / to zero. If every element in the main diagonal
is zero and b tj ?± 0, then adding row j to row / and column j to column /*
will yield a matrix with 2b 4j in the rth place of the main diagonal. The method
is a little fussy because the same row and column operations must be used,
and in the same order.
Another good method for quadratic forms of low order is called com
pleting the square. If X T BX = 2"j=i x &a x i anc * b u ^ 0> then
X T BX  f (b a x 1 + • • • + b in x n f
(10.3)
10 I The Normal Form
167
is a quadratic form in which x i does not appear. Make the substitution
x'i = b n x 1 + • • • + b in x n .
(10.4)
Continue in this manner if possible. The steps must be modified if at any
stage every element in the main diagonal is zero. If b tj ^ 0, then the sub
stitution x'. = x t + x t and x'. = x i  x t will yield a quadratic form repre
sented by a matrix with 7b u in the rth place of the main diagonal and lb „
in the y'th place. Then we can proceed as before. In the end we will have
x T Bx = f (*;)■ + ■■
Oil
(10.5)
expressed as a sum of squares; that is, the quadratic form will be in diagonal
form.
The method of elementary row and column operations and the method
of completing the square have the advantage of being based on concepts
much less sophisticated than the linear functional. However, the com
putational method based on the proof of the theorem is shorter, faster,
and more compact. It has the additional advantage of giving the matrix
of transition without special effort.
EXERCISES
1. Reduce each of the following symmetric matrices to diagonal form. Use the
method of linear functionals, the method of elementary row and column operations,
and the method of completing the square,
(«)
(c)
2 2"
1 2
2 2
1
1
2
1
1
1
1
1
1
1
2"
1
1
(*)
id)
1 2
2
3 1
3"
1
1
'0 12 3'
10 12
2 10 1
3 2 10
2. Using the methods of this section, reduce the quadratic forms of Exercise 1,
Section 9, to diagonal form.
3. Each of the quadratic forms considered in Exercise 2 has integral coefficients.
Obtain for each a diagonal form in which each coefficient in the main diagonal is
a squarefree integer.
168 Linear Functional, Bilinear Forms, Quadratic Forms  IV
11 I Real Quadratic Forms
A quadratic form over the complex numbers is not really very interesting.
From Theorem 10.2 we see that two different quadratic forms would be
distinguishable if and only if they had different ranks. Two quadratic forms
of the same rank each have coordinate systems (very likely a different
coordinate system for each) in which their representations are the same.
Hence, any properties they might have which would be independent of the
coordinate system would be indistinguishable.
In this section let us restrict our attention to quadratic forms over the
field of real numbers. In this case, not every number is a square; for
example, —1 is not a square. Therefore, having obtained a diagonalized
representation of a quadratic form, we cannot effect a further transformation,
as we did in the proof of Theorem 10.2 to obtain all l's for the nonzero
elements of the main diagonal. The best we can do is to change the positive
elements to + l's and the negative elements to — l's. There are many choices
for a basis with respect to which the representation of the quadratic form has
only +l's and —l's along the main diagonal. We wish to show that the
number of + l's and the number of — l's are independent of the choice of the
basis ; that is, these numbers are basic properties of the underlying quadratic
form and not peculiarities of the representing matrix.
Theorem 11.1. Let q be a quadratic form over the real numbers. Let P be
the number of positive terms in a diagonalized representation of q and let N
be the number of negative terms. In any other diagonalized representation of
q the number of positive terms is P and the number of negative terms is N.
proof. Let A = {a l5 . . . , a n } be a basis which yields a diagonalized
representation of q with P positive terms and N negative terms in the main
diagonal. Without loss of generality we can assume that the first P elements
of the main diagonal are positive. Let B = {/J l5 ...,/?„} be another basis
yielding a diagonalized representation of q with the first P' elements of the
main diagonal positive.
Let U = (a l5 . . . , a P > and let W = P > +1 , . . . , n ). Because of the
form of the representation using the basis A, for any nonzero a e U we have
q(a.) > 0. Similarly, for any peW we have q(jl) < 0. This shows that
U nW = {0}. Now dim U = P, dim W = n  P' , and dim (U + W) < n.
Thus P + n  P' = dim U + dim W = dim (U + W) + dim (U n W) =
dim (U + W) < n. Hence, P — P' < 0. In the same way it can be shown
that P'  P < 0. Thus P = P' and N = r  P =  P' = N'. □
Definition. The number S = P — N is called the signature of the quadratic
form q. Theorem 11.1 shows that S is well defined. A quadratic form is
called nonnegative semidefinite if S = r. It is called positive definite if S = n.
169
1 1  Real Quadratic Forms
It is clear that a quadratic form is nonnegative semidefinite if and only
if q(a) > for all a e V. It is positive definite if and only if q(a) > for
nonzero a e V. These are the properties of nonnegative semidefinite
and positive definite forms that make them of interest. We use them ex
tensively in Chapter V.
If the field of constants is a subfield of the real numbers, but not the
real numbers, we may not always be able to obtain +l's and l's along
the main diagonal of a diagonalized representation of a quadratic form.
However, the statement of Theorem 11.1 and its proof referred only to the
diagonal terms as being positive or negative, not necessarily +1 or 1.
Thus the theorem is equally valid in a subfield of the real numbers, and the
definitions of the signature, nonnegative semidefiniteness, and positive
definiteness have meaning.
In calculus it is shown that
I
e~ xt dx= 7T Vi .
It happens that analogous integrals of the form
C oo /*oo
J = • • • gS*ia«*, dx x  ■ • dx n
J— 00 J — 00
appear in a number of applications. The term J x fia x i = xT AX appearing
in the exponent is a quadratic form, and we can assume it to be symmetric.
In order that the integrals converge it is necessary and sufficient that the
quadratic form be positive definite. There is a nonsingular matrix P such
that P T AP = L is a diagonal matrix. Let {X lt . . . , K) be the main diagonal
of L. If X = (aj 1} ...,»„} are the old coordinates of a point, then Y =
NT . $ X i _
(2/i, • • • , y n ) are the new coordinates where x t = ZiP^y Since ^7 — Pw
the Jacobian of the coordinate transformation is det P. Thus, °
f 00 /*oo
= ... e ^iv<* det Pd Vl  dy n
J—ao J— 00
= det P\ e'^ yi2 dy x \ e~ XnVn dy n
J— 00 J°°
= deti > '"&
«/e det P
= 7r ^■■■X^'
170 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
Since X 1 • • ■ X n = det L = det P det A det P = det P 2 det A, we have
it/2
I =
det A"
EXERCISES
1. Determine the rank and signature of each of the quadratic forms of Exercise 1,
Section 9.
2. Show that the quadratic form Q(x, y) = ax 2 + bxy + cy\a, b, c real) is
positive definite if and only if a > and b 2 — Aac < 0.
3. Show that if A is a real symmetric positive definite matrix, then there exists
a real nonsingular matrix P such that A = P T P.
4. Show that if A is a real nonsingular matrix, then A T A is positive definite.
5. Show that if A is a real symmetric nonnegative semidefinite matrix — that is,
A represents a nonnegative semidefinite quadratic form — then there exists a
real matrix R such that A = R T R.
6. Show that if A is real, then A T A is nonnegative semidefinite.
7. Show that if A is real and A T A = 0, then A = 0.
8. Show that if A is real symmetric and A 2 = 0, then A = 0.
9. If /* 1? . . . , A r are real symmetric matrices, show that
A 2 + ■ ■ ■ + A 2 =
implies A 1 = A 2 — • • • = A r = 0.
12 I Hermitian Forms
For the applications of forms to many problems, it turns out that a
quadratic form obtained from a bilinear form over the complex numbers
is not the most useful generalization of the concept of a quadratic form
over the real numbers. As we see later, the property that a quadratic form
over the real numbers be positivedefinite is a very useful property. While
x 2 is positivedefinite for real x, it is not positivedefinite for complex x.
When dealing with complex numbers we need a function like a; 2 = xx,
where x is the conjugate complex of x. xx is nonnegative for all complex
(and real) x, and it is zero only when x = 0. Thus xx is a form which has
the property of being positive definite. In the spirit of these considerations,
the following definition is appropriate.
Definition. Let F be the field of complex numbers, or a subfield of the
complex numbers, and let V be a vector space over F. A scalar valued
12  Hermitian Forms 171
function /of two vectors, a, /? £ V is called a Hermitian form if
(1) /(a, /?)=/(£, a). (12.1)
(2) /(a, Z>A + £ 2 &) = V("> ft) + V(a, A)
A Hermitian form differs from a symmetric bilinear form in the taking
of the conjugate complex when the roles of the vectors <x and @ are inter
changed. But the appearance of the conjugate complex also affects the
bilinearity of the form. Namely,
/(fliai + a 2 a 2 , ft) =/(/?, a x v. x + a 2 a 2 )
= *i/(/0, ai) + aafiP, «2>
= aJX/?, ax) + fla/Off, a 2 )
= flj/Cax, ]8) + a 2 /(a 2 , /?).
We describe this situation by saying that a Hermitian form is linear in the
second variable and conjugate linear in the first variable.
Accordingly, it is also convenient to define a more appropriate general
ization to vector spaces over the complex numbers of the concept of a
bilinear form on vector spaces over the real numbers. A function of two
vectors on a vector space over the complex numbers is said to be conjugate
bilinear if it is conjugate linear in the first variable and linear in the second.
We say that a function of two vectors is Hermitian symmetric if /(a, /S) =
/(/?, a). This is the most useful generalization to vector spaces over the
complex numbers of the concept of symmetry for vector spaces over the
real numbers. In this terminology a Hermitian form is a Hermitian sym
metric conjugate bilinear form.
For a given Hermitian form/, we define q(<x) =/(a, a) and obtain what
we call a Hermitian quadratic form. In dealing with vector spaces over the
field of complex numbers we almost never meet a quadratic form obtained
from a bilinear form. The useful quadratic forms are the Hermitian quadratic
forms.
Let A = {a l5 . . . , a n } be any basis of V. Then we can let/(a i , a,) = h u
and obtain the matrix H = [h tj ] representing the Hermitian form / with
respect to A. H has the property that h it =f(ct it a.,) =/(a 3 , a,) = h H ,
and any matrix which has this property can be used to define a Hermitian
form. Any matrix with this property is called a Hermitian matrix.
If A is any matrix, we denote by A the matrix obtained by taking the
conjugate complex of every element of A; that is, if A = [a^thenA = [a u ].
We denote A T = A T by A*. In this notation a matrix His Hermitian if and
only if H* = H.
If a new basis 8 = {/? 1} . . . , @ n } is selected, we obtain the representation
172
Linear Functionals, Bilinear Forms, Quadratic Forms  IV
H' = [h'^] where h' tJ = f(fa, fa). Let P be the matrix of transition; that
is > fa = SU Pa** Then
K t =m,h)
(n n \
2 Pki*k, 2 PsM
n in
= 2P«i2 Pkif(*k,*s)
s =l fc=l
« n
= Z, .2, PkihjcsPsj
(12.3)
In matrix form this equation becomes //' = P*HP.
Definition. If a nonsingular matrix P exists such that H' = P*HP, we say
that H and H' are Hermitian congruent.
Theorem 12.1. For a given Hermitian matrix H there is a nonsingular
matrix P such that H' = P*HP is a diagonal matrix. In other words, iff is
the underlying Hermitian form, there is basis A' = {a[, . . . , a^} such that
/(<x t ', a J) = whenever i ^ j.
proof. The proof is almost identical with the proof of Theorem 10. 1 , the
corresponding theorem for bilinear forms. There is but one place where
a modification must be made. In the proof of Theorem 10.1 we made use of
a formula for recovering the symmetric part of a bilinear form from the
associated quadratic form. For Hermitian forms the corresponding formula
is
±[?(a + fa  q{* fa iq(* + ifa + ty(a  ifa) = /(«> fa ( l2A )
Hence, if /is not identically zero, there is an o^ e V such that qfa) ^ 0.
The rest of the proof of Theorem 10.1 then applies without change. □
Again, the elements of the diagonal matrix thus obtained are not unique.
We can transform H' into still another diagonal matrix by means of a
diagonal matrix Q with x x , . . . , x n , x i ^ 0, along the main diagonal. In this
fashion we obtain _
H" = Q*H'Q =
(12.5)
12 I Hermitian Forms
173
We see that, even though we are dealing with complex numbers, this trans
formation multiplies the elements along the main diagonal of H' by positive
real numbers.
Since q(«.) = /(a, a) = /(a, a), q{ct) is always real. We can, in fact, apply
without change the discussion we gave for the real quadratic forms. Let
P denote the number of positive terms in the diagonal representation of q,
and let N denote the number of negative terms in the main diagonal. The
number S = P — N is called the signature of the Hermitian quadratic
form q. Again, P + N = r, the rank of q.
The proof that the signature of a Hermitian quadratic form is independent
of the particular diagonalized representation is identical with the proof
given for real quadratic forms.
A Hermitian quadratic form is called nonnegative semidefinite if S = r.
It is called positive definite if S = n. If/is a Hermitian form whose associated
Hermitian quadratic form q is positivedefinite (nonnegative semidefinite),
we say that the Hermitian form / is positivedefinite (nonnegative semi
definite).
A Hermitian matrix can be reduced to diagonal form by a method analo
gous to the method described in Section 10, as is shown by the proof of
Theorem 12.1. A modification must be made because the associated Her
mitian form is not bilinear, but complex bilinear.
Let a( be a vector for which q{a£ ^ 0. With this fixed &[, f (a.^ a) defines
a linear functional <j>[ on V. If 04 is represented by
On, • • • , pm) = p and a b y ( x i> ...,xj = x,
then
n n
/(ai, a) = ^ 2 TTxKi x i
i=i i=i
n / n \
= l(lp7.h i) )x, (12.6)
This means the linear functional <j>[ is represented by P*H.
EXERCISES
1. Reduce the following Hermitian matrices to diagonal form.
(a)
"1 f
1 1
(b)
 1 1  r
1 +/
1
2. Let/be an arbitrary complex bilinear form. Define/* by the rule,/*(a, /?) =
/(/3, a). Show that/* is complex bilinear.
174 Linear Functionals, Bilinear Forms, Quadratic Forms  IV
3. Show that if H is a positive definite Hermitian matrix — that is, H represents
a positive definite Hermitian form — then there exists a nonsingular matrix P such
that H = P*P.
4. Show that if A is a complex nonsingular matrix, then A* A is a positive
definite Hermitian matrix.
5. Show that if H is a Hermitian nonnegative semidefinite matrix — that is, H
represents a nonnegative semidefinite Hermitian quadratic form — then there
exists a complex matrix R such that H = R*R.
6. Show that if A is complex, then A* A is Hermitian nonnegative semidefinite.
7. Show that if A is complex and A* A = 0, then A = 0.
8. Show that if A is hermitian and A 2 = 0, then A = 0.
9. If A x , . . . , A r are Hermitian matrices, show that A x 2 + • • ■ + A* =
implies A x = • • • = A r = 0.
10. Show by an example that, if A and B are Hermitian, it is not necessarily
true that AB is Hermitian. What is true if A and B are Hermitian and AB = BA1
chapter
V
Orthogonal
and unitary
transformations,
normal matrices
In this chapter we introduce an inner product based on an arbitrary positive
definite symmetric bilinear form, or Hermitian form. On this basis the
length of a vector and the concept of orthogonality can be defined. From
this point on, we concentrate our attention on bases in which the vectors are
mutually orthogonal and each is of length 1 , the orthonormal bases. The
GramSchmidt process for obtaining an orthonormal basis from an arbitrary
basis is described.
Isometries are linear transformations which preserve length. They also
preserve the inner product and therefore map orthonormal bases onto
orthonormal bases. It is shown that a matrix representing an isometry has
exactly the same properties as a matrix of transition representing a change
of bases from one orthonormal basis to another. If the field of scalars is
real, these matrices are said to be orthogonal; and if the field of scalars is
complex, they are said to be unitary.
If A is an orthogonal matrix, we show that A T = A 1 ; and if A is unitary,
we show that A* = A 1 . Because of this fact a matrix representing a linear
transformation and a matrix representing a bilinear form are transformed
by exactly the same formula under a change of coordinates provided that
the change is from one orthonormal basis to another. This observation
unifies the discussions of Chapter III and IV.
The penalty for restricting our attention to orthonormal bases is that
there is a corresponding restriction in the linear transformations and bilinear
forms that can be represented by diagonal matrices. The necessary and
sufficient condition that this be possible, expressed in terms of matrices,
is that A* A = A A*. Matrices with this property are called normal matrices.
Fortunately, the normal matrices constitute a large class of matrices and
175
176 Orthogonal and Unitary Transformations, Normal Matrices  V
they happen to include as special cases most of the types that arise in physical
problems.
Up to a certain point we can consider matrices with real coefficients to
be special cases of matrices with complex coefficients. However, if we wish
to restrict our attention to real vector spaces, then the matrices of transition
must also be real. This restriction means that the situation for real vector
spaces is not a special case of the situation for complex vector spaces. In
particular, there are real normal matrices that are unitary similar to diagonal
matrices but not orthogonal similar to diagonal matrices. The necessary
and sufficient condition that a real matrix be orthogonal similar to a diagonal
matrix is that it be symmetric.
The techniques for finding the diagonal normal form of a normal matrix
and the unitary or orthogonal matrix of transition are, for the most part,
not new. The eigenvalues and eigenvectors are found as in Chapter III. We
show that eigenvectors corresponding to different eigenvalues are automati
cally orthogonal so all that needs to be done is to make sure that they are of
length 1. However, something more must be done in the case of multiple
eigenvalues. We are assured that there are enough eigenvectors, but we
must make sure they are orthogonal. The GramSchmidt process provides
the method for finding the necessary orthohormal eigenvectors.
1 I Inner Products and Orthogonal Bases
Even when speaking in abstract terms we have tried to draw an analogy
between vector spaces and the geometric spaces we have encountered in
2 and 3dimensional analytic geometry. For example, we have referred to
lines and planes through the origin as subspaces ; however, we have nowhere
used the concept of distance. Some of the most interesting properties of
vector spaces and matrices deal with the concept of distance. So in this
chapter we introduce the concept of distance and explore the related proper
ties.
For aesthetic reasons, and to show as clearly as possible that we need not
have an a priori concept of distance, we use an approach which will emphasize
the arbitrary nature of the concept of distance.
It is customary to restrict attention to the field of real numbers or the field
of complex numbers when discussing vector space concepts related to dis
tance. However, we need not be quite that restrictive. The scalar field F
must be a subfield of the complex numbers with the property that, if a e F,
the conjugate complex a is also in F. Such a field is said to be normal over its
real subfield. The real field and the complex field have this property, but
so do many other fields. For most of the important applications of the mate
rial to follow the field of scalars is taken to be the real numbers or the field
1 I Inner Products and Orthogonal Bases 177
of complex numbers. Although most of the proofs given will be valid for
any field normal over its real subfield, it will suffice to think in terms of the
two most important cases.
In a vector space V of dimension n over the complex numbers (or a subfield
of the complex numbers normal over its real subfield), let /be any fixed
positive definite Hermitian form. For the purpose of the following develop
ment it does not matter which positive definite Hermitian form is chosen,
but it will remain fixed for all the remaining discussion. Since this particular
Hermitian form is now fixed, we write (a, /?) instead of /(a, /?). (a, /?) is
called the inner product , or scalar product, of a and /5.
Since we have chosen a positive d efinite Hermitian form, (a, a) >
and (a, a) > unless a = 0. Thus V(a, a) = a is a welldefined non
negative r eal num ber which we call the length or norm of a. Observe that
11**11 = v(tfa, aa) = \laa(<x, a) = \a\ • a, so that multiplying a vector
by a scalar a multiplies its length by \a\. We say that the distance between
two vectors is the norm of their difference; that is, d(&, p) = \\0  <x.
We should like to show that this distance function has the properties we
might reasonably expect a distance function to have. But first we have to
prove a theorem that has interest of its own and many applications.
Theorem 1.1. For any vectors a,^V, (a, 0)\ < a • g. This in
equality is known as Schwarz's inequality.
proof. For t a real number consider the inequality
< (*, fit*  fl« = (a, 0)» «[* t*  It  (a, P)\ 2 + /?«. (1.1)
If  a  = 0, the fact that this inequality must hold for arbitrarily large t
implies that (a, f})\ = so that Schwarz's inequality is satisfied. If a ^ 0,
take t = 1/ 1 a 2 . Then (1.1) is equivalent to Schwarz's inequality,
l(*,/?)l< lal/?ll.n (1.2)
This proof of Schwarz's inequality does not make use of the assumption
that the inner product is positive definite and would remain valid if the
inner product were merely semidefinite. Using the assumption that the
inner product is positive definite, however, an examination of this proof of
Schwarz's inequality would reveal that equality can hold if and only if
(*(,*■ = *> (13)
(a, a)
that is, if and only if /? is a multiple of a.
[f a ?£ and ^ 0, Schwarz's inequality can be written in the form
7^ < 1. (1.4)
178 Orthogonal and Unitary Transformations, Normal Matrices  V
In vector analysis the scalar product of two vectors is equal to the product
of the lengths of the vectors times the cosine of the angle between them.
The inequality (1.4) says, in effect, that in a vector space over the real
numbers the ratio ' can be considered to be a cosine. It would be
11*11 • IIpII
a diversion for us to push this point much further. We do, however, wish
to show that d(oc, (S) behaves like a distance function.
Theorem 1.2. For d(a, fi) = \\(t — a, we have,
(1) rf(a, 0) = d(P, a),
(2) </(a, p) > and d(a, 0) = if and only ifot = fi,
(3) d(x, 0) < d(a, y) + d(y, ft.
proof. (1) and (2) are obvious. (3) follows from Schwarz's inequality.
To see this, observe that
a + /? 2 = (a + 0,a + /0)
= (a, a) + (a, 0) + (0, a) + (P, j8)
= a 2 + (a, / S) + (a,£) + /?«
<  a  2 + 2 (a,0) + II^P
< Ml 2 + 2 a • W\\ + \\PV = (11*11 + WWf (1.5)
Replacing a by y — a and /? by /? — y, we have
11/8 a < ya + ll£y.n (1.6)
(3) is the familiar triangular inequality. It implies that the sum of two small
vectors is also small. Schwarz's inequality tells us that the inner product
of two small vectors is small. Both of these inequalities are very useful for
these reasons.
According to Theorem 12.1 of Chapter IV and the definition of a positive
definite Hermitian form, there exists a basis A = {a 1? . . . , a w } with respect
to which the representing matrix is the unit matrix. Thus,
(a„ a,) = <5„. (1.7)
Relative to this fixed positive definite Hermitian form, the inner product,
every set of vectors that has this property is called an orthonormal set.
The word "orthonormal" is a combination of the words "orthogonal"
and "normal." Two vectors a and fi are said to be orthogonal if (a, /9) =
(jS, a) = 0. A vector a is normalized if it is of length 1 ; that is, if (a, a) =
1. Thus the vectors of an orthonormal set are mutually orthogonal and nor
malized. The basis A chosen above is an orthonormal basis. We shall see
that orthonormal bases possess particular advantages for dealing with the
properties of a vector space with an inner product. A vector space over the
complex numbers with an inner product such as we have defined is called
1 I Inner Products and Orthogonal Bases 179
a unitary space. A vector space over the real numbers with an inner product
is called a Euclidean space.
For <x, e V, let a = 2? =1 s 4 a, and /ff = 2? =1 y<a f . Then
= 2
X,
2 y^** a i)
t=i i_j=i
= 2 x iVi . (1.8)
«=i
If we represent a by the ntuple (x lf . . . , scj = X, and /? by the ntuple
(y 1? . . . , y n ) = Y, the inner product can be written in the form
(a,/0 = i^ = x*y. (19)
i=l
This is a familiar formula in vector analysis where it is also known as the
inner product, scalar, or dot product.
Theorem 1.3. An orthonormal set is linearly independent.
proof. Suppose that {£ l5 £ 2 , . . . } is an orthonormal set and that ^i x£i =
0. Then = (£,, 0) = (£,, f t *,&) = 2, *<(* y. W = x i Thus the set is
linearly independent. □
It is an immediate consequence of Theorem 1.3 that an orthonormal set
cannot contain more than n elements.
Since V has at least one orthonormal basis and orthonormal sets are
linearly independent, some questions naturally arise. Are there other
orthonormal bases ? Can an orthonormal set be extended to an orthonormal
basis ? Can a linearly independent set be modified to form an orthonormal
set? For infinite dimensional vector spaces the question of the existence
of even one orthonormal basis is a nontrivial question. For finite dimen
sional vector spaces all these questions have nice answers, and the technique
employed in giving these answers is of importance in infinite dimensional
vector spaces as well.
Theorem 1.4. IfA = {a l9 . . . , a s } is any linearly independent set whatever
in V, there exists an orthonormal set X = { l5 . . . ,  s } such that £ fe = 2i=i a ifc a i
proof. (The GramSchmidt orthonormalization process). Since a x is
an element of a linearly independent set <x x ^ 0, and therefore HaJ > 0.
Let Z 1 = T r. oci Clearly, U x \\ = 1.
IK II
Suppose, then, {^, . . . , £ r } has been found so that it is an orthonormal
set and such that each £ k is a linear combination of {a l5 . . . , a^.}. Let
<*Ul = a r+l  (fl» <*r4l)fl  '  '  (£■> <*r+l)£r C 1 ' 10 )
180 Orthogonal and Unitary Transformations, Normal Matrices  V
Then for any £„ 1 < i < r, we have
(ft, < + i) = (ft> O  (ft> a r + i) = 0. (1.11)
Furthermore, since each  fc is a linear combination of the {a x , . . . , aj,
a; +l is a linear combination of the {a l5 . . . , a r+1 }. Also, < +1 is not zero
since {a x , . . . , a r+1 } is a linearly independent set and the coefficient of
a r+1 in the representation of a' r+1 is 1. Thus we can define
II Or +1 II
Clearly, { l5 . . . , £ r+1 } is an orthonormal set with the desired properties.
We can continue in this fashion until we exhaust the elements of A. The
set X = {g lt . . . , J has the required properties. □
The GramSchmidt process is completely effective and the computations
can be carried out exactly as they are given in the proof of Theorem 1.4.
For example, let A = {a x = (1, 1,0, 1), a 2 = (3, 1, 1, 1), a 3 = (0, 1, 1,
1)}. Then
£ 1 = ^=(1,1,0,1),
V3
a 2 = (3, 1, 1, 1)  4=4=(1, 1,0, 1) = (2,0, 1, 2),
V3V3
£ 2 = 1(2, 0, 1, 2),
a 3 = (0, l, l, l)  4~7= (1 ' i  1} ~ l^ 2 ' ' u ~ 2)
V3V3 3 3
= 1(0,1,2,1),
 3 = i(0, 1,2,1).
V6
It is easily verified that {&, £ 2 , f 8 > is an orthonormal set.
Corollary 1.5. If A = {a l5 . . . , aj is a basis of V, the orthonormal set
X = {f x , . . . , J, obtained from A by the application of the GramSchmidt
process, is an orthonormal basis of V.
proof. Since X is orthonormal it is linearly independent. Since it
contains n vectors it also spans V and is a basis. □
Theorem 1.4 and its corollary are used in much the same fashion in which
we used Theorem 3.6 of Chapter I to obtain a basis (in this case an ortho
normal basis) such that a subset spans a given subspace.
1 I Inner Products and Orthogonal Bases 181
Theorem 1.6. Given any vector a x of length 1, there is an orthonormal
basis with a x as the first element.
proof. Since the set {o^} is linearly independent it can be extended to a
basis with a x as the first element. Now, when the GramSchmidt process
is applied, the first vector, being of length 1, is unchanged and becomes the
first vector of an orthonormal basis. □
EXERCISES
In the following problems we assume that all ntuples are representations of
their vectors with respect to orthonormal bases.
1. Let A ={<*!,..., a 4 } be an orthonormal basis of R 4 and let a, fie V be
represented by (1, 2, 3, 1) and (2, 4,1, 1), respectively. Compute (a, p).
2. Let a = (1, /, 1 + /) and P = (/, 1, / — 1) be vectors in C 3 , where C is the
field of complex numbers. Compute (a, p).
3. Show that the set {(1, /, 2), (1, i, 1), (1, /, 0)} is orthogonal in C 3 .
4. Show that (a, 0) = (0, a) = for all a £ V.
5. Show that a + pf + <x  /? 2 = 2 oc 2 + 2 £ 2 .
6. Show that if the field of scalars is real and a = \\p\\, then a  p and <x + p
are orthogonal, and conversely.
7. Show that if the field of scalars is real and a + p\\ 2 = a 2 + £ 2 , then
a and P are orthogonal, and conversely.
8. Verify Schwarz's inequality for the vectors a and jS in Exercises 1 and 2.
9. The set {(1, 1, 1), (2, 0, 1), (0, 1, 1)} is linearly independent, and hence a
basis for F 3 . Apply the GramSchmidt process to obtain an orthonormal basis.
10. Given the basis {(1,0, 1, 0), (1, 1, 0, 0), (0, 1, 1, 1,), (0, 1,1,0)} apply the
GramSchmidt process to obtain an orthonormal basis.
11. Let W be a subspace of V spanned by {(0, 1,1,0), (0, 5, 3, 2), (3,
—3, 5, —7)}. Find an orthonormal basis for W.
12. In the space of real integrable functions let the inner product be defined by
«i
f(x)g(x) dx.

Find a polynomial of degree 2 orthogonal to 1 and x. Find a polynomial of degree
3 orthogonal to 1, x, and x 2 . Are these two polynomials orthogonal?
13. Let X = {£j, . . . , £ TO } be a set of vectors in the ndimensional space V.
Consider the matrix G — [gjj] where
ga = (ft. £*)•
Show that if X is linearly dependent, then the columns of G are also linearly
dependent. Show that if X is linearly independent, then the columns of G are also
182 Orthogonal and Unitary Transformations, Normal Matrices  V
linearly independent. Det G is known as the Gramian of the set X. Show that X is
linearly dependent if and only if det G = 0. Choose an orthonormal basis in V and
represent the vectors in X with respect to that basis. Show that G can be represented
as the product of an m x n matrix and an n x m matrix. Show that det G > 0.
*2 I Complete Orthonormal Sets
We now develop some properties of orthonormal sets that hold in both
finite and infinite dimensional vector spaces. These properties are deep
and important in infinite dimensional vector spaces, but in finite dimensional
vector spaces they could easily be developed in passing and without special
terminology. It is of some interest, however, to borrow the terminology of
infinite dimensional vector spaces and to give proofs, where possible, which
are valid in infinite as well as finite dimensional vector spaces.
Let X = {£ l5  2 , . . . } be an orthonormal set and let a be any vector in V.
The numbers {o^ = ( i5 a)} are called the Fourier coefficients of a.
There is, first, the question of whether an expression like 2* x £i nas anv
meaning in cases where infinitely many of the x i are nonzero. This is a
question of the convergence of an infinite series and the problem varies
from case to case so that we cannot hope to deal with it in all generality.
We have to assume for this discussion that all expressions like 2* x Si tnat
we write down have meaning.
Theorem 2.1. The minimum of a — 2* x i£A\ * s attained if and only if all
x i = (&, x ) = «<•
PROOF.
II*  2 x Sif = (a  2 x £i> a  2 x &)
i i i
= (a, a)  2 X A  2 x i a i + 2 x i x i
i i i
= 1 afii  ^ X A  2 x i a i + 2 x i x i + ( a > a )  2 d i a i
i i i i i
= 2 («.■  x i)( a i  x i) + II a f  2 *i°i
i i
= 2 k  x i\ 2 + ll«ll"  2 kl 2  (2.D
i i
Only the term 2* \ a i ~ x i\ 2 depends on the x t and, being a sum of real
squares, it takes on its minimum value of zero if and only if all x t = a t . □
Theorem 2.1 is valid for any orthonormal set X, whether it is a basis or
not. If the norm is used as a criterion of smallness, then the theorem says
that the best approximation of a in the form 2* x £i (using only the £ f e X)
is obtained if and only if all x i are the Fourier coefficients.
2  Complete Orthonormal Sets * 83
Theorem 2.2 2* k 2 < a 2 . This inequality is known as BesseVs in
equality.
proof. Setting x t = a t in equation (2.1) we have
la 2 2l^ 2 =ll a I«^ll 2 ^° D (2,2)
It is desirable to know conditions under which the Fourier coefficients
will represent the vector a. This means we would like to have a = 2t a*!*.
In a finite dimensional vector space the most convenient sufficient con
dition is that X be an orthonormal basis. In the theory of Fourier series
and other orthogonal functions it is generally not possible to establish
the validity of an equation like a = 2* «,£< without some modification of
what is meant by convergence or a restriction on the set of functions under
consideration. Instead, we usually establish a condition known as com
pleteness. An orthonormal set is said to be complete if and only if it is not
a subset of a larger orthonormal set.
Theorem 2.3. Let X = {£ <} be an orthonormal set. The following three
conditions are equivalent:
(1) For each aJeV, (a, £) = 2 (£Ta)(£„ 0). ( 2  3 )
i
(2) For each a e V, a 2 = 2 1(1,, a) 2 . (24)
(3) X is complete.
Equations (2.3) and (2.4) are both known as Parseva Vs ide ntities.
proof. Assume (1). Then a 2 = (a, a) = 2* („ a)(£„ a)
Assume (2). If X were not complete, it would be contained in a larger
orthonormal set Y. But for any a eY,aL $ X, we would have
l=laol 2 =2K^«o) 2 =
i
because of (2) and the assumption that Y is orthonormal. Thus X is complete.
Now, assume (3). Let be any vector in V and consider /?' = p — 2* (!<,
/%. Then
(f,,W«(*«,/82(^0)£i)
* (ft,  2 (f, .#(****)
= (£,/?) (£,/?) =0;
that is, 0' is orthogonal to all £, e X. If 0' * 0, then X U pj £'}
184 Orthogonal and Unitary Transformations, Normal Matrices  V
would be a larger orthonormal set. Hence, \\@'\\ = 0. Using the assumption
that the inner product is positive definite we can now conclude that ft' = 0.
However, it is not necessary to use this assumption and we prefer to avoid
using it. What we really need to conclude is that if a is any vector in V then
(a, /?') = 0, and this follows from Schwarz's inequality. Thus we have
i
= (",£) 2 (^)(a,l*)
i
or
(a, = 2 (**«)(&#■
i
This completes the cycle of implications and proves that conditions (1),
(2), and (3) are equivalent. □
Theorem 2.4. The following two conditions are equivalent:
(4) The only vector orthogonal to all vectors in X is the zero vector.
(5) For each a e V, a = 2 (f*» *)&• ( 2 5)
i
proof. Assume (4). Let a be any vector in V and consider a' = a —
2^ (I*, a)£ 4 . Then
(£„«') = (£«,« 2 (**«)*<)
= &,«) 2 (**.«)&.**)
= (£, a)  (£, a) = 0;
that is, a' is orthogonal to all ^ g X. Thus a' = and a = 2» (£»•> a )£i
Now, assume (5) and let a be orthogonal to all £ t g X. Then a =
2, (*„ a)*< = 0. a
Theorem 2.5. The conditions (4) or (5) wn/?/y ?/ze conditions (1), (2), and
(3).
proof. Assume (5). Then
(a, = (2 (£.«)£«. 2 (**/%)
= 2(ft.«)2(^0(^W
= 2 (^ «)(£<> 0 □
i
Theorem 2.6. If the inner product is positive definite, the conditions (1),
(2), or (3) imply the conditions (4) and (5).
2  Complete Orthonormal Sets 185
proof. In the proof that (3) implies (1) we showed that if a' = a 
2< (fi, a)!,, then a' = 0. If the inner product is positive definite, then
a' = and, hence,
a = J ( if a),. D
The proofs of Theorems 2.3, 2.4, and 2.5 did not make use of the positive
definiteness of the inner product and they remain valid if the inner product
is merely nonnegative semidefinite. Theorem 2.6 depends critically on
the fact that the inner product is positive definite.
For finite dimensional vector spaces we always assume that the inner
product is positive definite so that the three conditions of Theorem 2.3 and
the two conditions of Theorem 2.4 are equivalent. The point of our making
a distinction between these two sets of conditions is that there are a number
of important inner products in infinite dimensional vector spaces that are
not positive definite. For example, the inner product that occurs in the
theory of Fourier series is of the form
( a , ft = 1 f * *@)P(x) dx. (2.6)
This inner product is nonnegative semidefinite, but not positive definite if
V is the set of integrable functions. Hence, we cannot pass from the com
pleteness of the set of orthogonal functions to a theorem about the con
vergence of a Fourier series to the function from which the Fourier
coefficients were obtained.
In using theorems of this type in infinite dimensional vector spaces in
general and Fourier series in particular, we proceed in the following manner.
We show that any a £ V can be approximated arbitrarily closely by finite
sums of the form J f xj t . For the theory of Fourier series this theorem is
known as the Weierstrass approximation theorem. A similar theorem must
be proved for other sets of orthogonal functions. This implies that the
minimum mentioned in Theorem 2.1 must be zero. This in turn implies that
condition (2) of Theorem 2.3 holds. Thus Parseval's equation, which is
equivalent to the completeness of an orthonormal set, is one of the principal
theorems of any theory of orthogonal functions. Condition (5), which is the
convergence of a Fourier series to the function which it represents, would
follow if the inner product were positive definite. Unfortunately, this is
usually not the case. To get the validity of condition (5) we must either add
further conditions or introduce a different type of convergence.
EXERCISES
1. Show that if X is an orthonormal basis of a finite dimensional vector space,
then condition (5) holds.
186 Orthogonal and Unitary Transformations, Normal Matrices  V
2. Let X be a finite set of mutually orthogonal vectors in V. Suppose that the
only vector orthogonal to each vector in X is the zero vector. Show that X is a
basis of V.
3 I The Representation of a Linear Functional by an Inner Product
For a fixed vector /? £ V, (/?, a) is a linear function of a. Thus there is a
linear functional e V such that 0(a) = (0, a) for all a. We denote the linear
functional defined in this way by <f> p . The following theorem is a converse
of this observation.
Theorem 3.1. Given a linear functional <f> e V, there exists a unique r\ g V
such that 0(a) = (rj, a.) for all a e V.
proof. Let X = {£ l5 . . . , £ n } be an orthonormal basis of V, and let
X = {(/>!,..., <f> n } be the dual basis. Let e V have the representation
4> = 2?=i Vih Define v = L n =i yJi Then for each %i> 0?> £i) =
(2?i $&> £) = 2r=i ytfi> **) = ^ = 2Ti y^<(^) = ^ (W But then
0(a) and (77, a) are both linear functional on V that coincide on the basis,
and hence coincide on all of V.
If r] x and t] 2 are two choices such that (r/ lt a) = (rj 2 , a) = 0(a) for all
a 6 V, then (^ — rj 2 , a) = for all a e V. For a = ^ — % this means
(?h — ??2> ?h — %) = ° Hence, r\ x — ?y 2 = and the choice for ?7 is
unique. □
Call the mapping denned by this theorem r\\ that is, for each e V,
rj(<f>)eV has the property that 0(a) = (^(0), a) for all a e V.
.A.
Theorem 3.2. The correspondence between <f> e V and ??(0) e V w onetoone
and onto V.
proof. In Theorem 3.1 we have already shown that r)(<f>) is well defined.
Let j8 be any vector in V and let <f> p be the linear functional in V such that
^(a) = (8, a) for all a. Then fi = f]((f> p ) and the mapping is onto. Since
(jft, a), as a function of a, determines a unique linear functional <f> p the
correspondence is onetoone. □
Theorem 3.3. If the inner product is symmetric, r\ is an isomorphism of V
onto V.
proof. We have already shown in Theorem 3.2 that rj is onetoone and
onto. Let = 2< M< and consider p = ]£< 6^(0*). Then (0, a) =
(2<M(&)> *) = (*,2<bM4>J) = I<b t (*, V(<f>i)) = Iibi(ri(<t>i), a) =
j, ^(a) = 0(a). Thus ??(0) = £ = £< bMh) and »? is linear  D
Notice that 77 is not linear if the scalar field is complex and the inner
product is Hermitian. Then for = ^ brfi we consider y = 2* hiVi^d
We see that (y, a) = (£, 5 ,»?(&), a ) = 2* &<(»?(&)> a ) = 2< *><&(<*) = <H a )
3  The Representation of a Linear Functional by an Inner Product 187
Thus rj(<f>) = y = 2* Birth) and r\ is conjugate linear. It should be observed
that even when r\ is conjugate linear it maps subspaces of V onto subspaces
of V.
We describe this situation by saying that we can "represent a linear func
tional by an inner product." Notice that although we made use of a particular
basis to specify the r\ corresponding to <f>, the uniqueness shows that this
choice is independent of the basis used. If V is a vector space over the real
numbers, <f> and v\ happen to have the same coordinates. This happy coin
cidence allows us to represent V in V and make V do double duty. This fact is
exploited in courses in vector analysis. In fact, it is customary to start
immediately with inner products in real vector spaces with orthonormal
bases and not to mention V at all. All is well as long as things remain simple.
As soon as things get a little more complicated, it is necessary to separate
the structure of V superimposed on V. The vectors representing themselves
in Vare said to be contravariant and the vectors representing linear functional
in V are said to be covariant.
We can see from the proof of Theorem 3.1 that, if V is a vector space
over the complex numbers, <f> and the corresponding r) will not necessarily
have the same coordinates. In fact, there is no choice of a basis for which
each <f> and its corresponding r\ will have the same coordinates.
Let us examine the situation when the basis chosen in V is not orthonormal.
Let A = {a l5 . . . , aj be any basis of V, and let A = {wi, • • • » V«} be the
corresponding dual basis of V. Let b it = (a„ a,). Since the inner product is
Hermitian, b ti = b H , or [b ti ] = B = B*. Since the inner product is positive
definite, B has rank n. That is, B is nonsingular. Let <f> = £r=i c t rp t be an
arbitrary linear functional in V. What are the coordinates of the correspond
ing r\1 Let r\ = ^U VPi Tnen
(V>*i) = (2y<«<> a *j
n
= 2 &( a *' a >)
= 2 vtK
n
= 2 c fc Vfc(a*)
fc=i
= c, (31)
Thus, we have to solve the equations
2 Ma = i Mi = c„ j = 1, . . . , n. (3.2)
i=l i=l
188 Orthogonal and Unitary Transformations, Normal Matrices  V
In matrix form this becomes
BY = C T ,
where
C = [c x • • • c n ],
or
Y = B~ X C* = (CB 1 )*. (3.3)
Of course this means that it is rather complicated to obtain the coordinate
representation of r\ from the coordinate representation of cf>. But that is
not the cause for all the fuss about covarient and contravariant vectors.
After all, we have shown that r\ corresponds to (J> independently of the basis
used and the coordinates of rj transform according to the same rules that
apply to any other vector in V. The real difficulty stems from the insistence
upon using (1.9) as the definition of the inner product, instead of using a
definition not based upon coordinates.
If V = 2?=i y< a i> and £ = 2"=i x i a i> we see that
(n n
= 2 ^Lvi b n x i
i=i j=i
= Y*BX. (3.4)
Thus, if r\ represents the linear functional <f>, we have
(rj, £) = Y*BX
= (CB'^BX
= CX
= (C*)*X. (3.5)
Elementary treatments of vector analysis prefer to use C* as the repre
sentation of rj. This preference is based on the desire to use (1.9) as the
definition of the inner product so that (3.5) is the representation of (rj, g),
rather than to use a coordinatefree definition which would lead to (rj, )
being represented by (3.4). The elements of C* are called the covariant
components of r\. We obtained C by representing </> in V. Since the dual space
is not available in such an elementary treatment, some kind of artifice must
be used. It is then customary to introduce a reciprocal basis A* = {a* , . . . , a*},
where a* has the property (a*, a,) = d^ = ^(a,). A* is the representation
of the dual basis A in V. But C was the original representation of <f> in
terms of the dual basis. Thus, the insistence upon representing linear
functionals by the inner product does not result in a single computational
advantage. The confusion that it introduces is a severe price to pay to avoid
introducing linear functionals and the dual space at the beginning.
4  The Adjoint Transformation *°9
4 I The Adjoint Transformation
Theorem 4.1. For a given linear transformation a on V, there is a unique
linear transformation a* on V such that (<r*(a), 0) = (a, o{fi))for all a, /? e V.
proof. Let a be given. Then for a fixed a, (a, <y(8)) is a linear function
of j8, that is, a linear functional on V. By Theorem 3.1 there is a unique // e V
such that (a, <r(#)) = (??, j8). Define a* (a) to be this 77.
Now, O^ + « 2 a 2 , (r(/8)) = a^, cr(#)) + a 2 (a 2 , o(p)) = a 1 (a*(a 1 ), /)) +
a 2 (a*(* 2 ), /?) = (fli<r*(a 1 ) + a 2 a*(a 2 ), /5) so that a 1 ff*(a 1 ) + a 2 cr*(a 2 ) =
a*{a 1 aL 1 + a 2 a 2 ) and or* is linear. □
Since for each <x the choice for <r*(a) is unique, cr* is uniquely defined by a.
a* is called the adjoint of cr.
Theorem 4.2. The relation between a and a* is symmetric, that is,
(o  *)* = a.
proof. Let a be given. Then cr* is defined uniquely by 0*(a), 0) =
(a, <r(0)) for all a, e V. Then (cr*)*, which we denote by a**, is defined
by (<x**(a), 0) = (a, cr*(/S)) for all a, e V. N ow the in ner product is
Hermitian so that (cr**(a), 0) = (a, <;*(/?)) = (cr*(£), a) = (0, (7(a)) =
((7(a), 0). Thus cr**(a) = cr(a) for all a e V; that is, cr** = cr. It then
follows also that (cr(o), p) = (a, cr *(#)). D
Let ^ = [a w ] be the matrix representing cr with respect to an orthonormal
basis X = {f f , . . . , J and let us find the matrix representing cr*.
(**(£,), &) = (£„ o(^))
i=l
n
= (i *«£,,£*)■ (4.D
Since this equation holds for all  fc , cr*(l ; ) = SU 5,^. Thus cr* is repre
sented by the conjugate transpose of A ; that is, a* is represented by A *.
The adjoint cr* is closely related to the dual a defined on page 142. a is a
linear transformation of V into itself, so the dual a is a linear transformation
of V into itself. Since r\ establishes a onetoone correspondence between
V and V, we can define a mapping of V into itself corresponding to a on
190 Orthogonal and Unitary Transformations, Normal Matrices  V
A.
V. For any a e V we can map a onto ^{o , [^~ 1 (a)]} and denote this map
ping by r](6). Then for any a, /? e V we have
= irHoOMfl]
= (a, a(fl)
= (or*(a), 0). (4.2)
Hence, ??(cr)(a) = cr*(a) for all a e V, that is, ^(cr) = cr*. The adjoint is a
representation of the dual. Because the mapping r\ of V onto V is conjugate
linear instead of linear, and because the vectors in V are represented by row
matrices while those in V are represented by columns, the matrix representing
cr* is the transpose of the complex conjugate of the matrix representing a.
Thus a* is represented by A*.
We shall maintain the distinction between the dual a defined on V and the
adjoint a* defined on V. This distinction is not always made and quite often
both terms are used for both purposes. Actually, this confusion seldom
causes any trouble. However, it can cause trouble when discussing the matrix
representation of a or cr*. If cr is represented by A, we have chosen also to
represent a by A with respect to the dual basis. If we had chosen to represent
linear functionals by columns instead of rows, a would have been repre
sented by A T . It would have been represented by A T in either the real or the
complex case. But the adjoint cr* is represented by A*. No convention will
allow a and cr* to be represented by the same matrix in the complex case
because the mapping r\ is conjugate linear. Because of this we have chosen to
make clear the distinction between a and cr*, even to the extent of having the
matrix representations look different. Furthermore, the use of rows to
represent linear functionals has the advantage of making some of the formulas
look simpler. However, this is purely a matter of choice and taste, and other
conventions, used consistently, would serve as well.
Since we now have a model of V in V, we can carry over into V all the
terminology and theorems on linear functionals in Chapter IV. In particular,
we see that an orthonormal basis can also be considered to be its own dual
basis since (f f , £,) = b ti .
Recall that, when a basis is changed in V and P is the matrix of transition,
(pTy\ i s ^e matrix of transition for the dual bases in V. In mapping V
onto V, {P T )~ X = (P*)' 1 becomes the matrix of transition for the representa
tion of dual basis in V. Since an orthonormal basis is dual to itself, if P is the
matrix of transition from one orthonormal basis to another, then P must also
be the matrix of transition for the dual basis; that is, (P*)~ x — P. This
important property of the matrices of transition from one orthonormal basis
to another will be established independently in Section 6.
1Q1
4  The Adjoint Transformation x * x
Let W be a subset of V. In Chapter IV4, we denned W± to be the anni
hilator of W in V. The mapping r\ of V onto V maps W^ onto a subspace
of V. It is easily seen that ??( W 1 ) is precisely the set of all vectors orthogonal
to every vector in W. Since we are in the process of dropping V as a separate
space and identifying it with V, we denote the set of all vectors in orthogonal
to all vectors in W by V/ 1  and call it the annihilator of W.
Theorem 4.3. If W is a subspace of dimension p, W x is of dimension
n  p . W n W^ = {0}. W@W± = V.
proof. That WJ is of dimension n  p follows from Theorem 4.1 of
Chapter IV. The other two assertions had no meaning in the context of
Chapter IV. IfaeWnW 1 , then Ha 2 = (a, a) = so that a = 0.
Since dim {W + W^} = dim W + dim WL  dim {W n W^} = p +
(„ _ p) = n , W ® W± = V. D
When Wi and W 2 are subspaces of V such that their sum is direct and W x
and W 2 are also orthogonal, we use the notation W 1 ± W 2 to denote this sum.
Actually, the fact that the sum is direct is a consequence of the fact that the
subspaces are orthogonal. In this notation, the direct sum in the con
clusion of Theorem 4.3 takes the form V = W _L W 1 .
Theorem 4.4. Let W be a subspace invariant under a. W x is then invariant
under o*.
proof. Let a e W ± . Then, for any e W, (<r*(a), 0) = (a, <r(0)) =
since <r(j8) £ W. Thus cr*(a) e WL. D
Theorem 4.5. ^(cr*) = Im(ff) 1 .
proof. By definition (a, otf)) = (<r*(a), (3). (a, a(/5)) = for all p e V
if and only if a e Im^ and (<r*(a), /?) = for all £ e V if and only if
a e #(<r*). Thus tf(or*) = lm(ay. a
Theorem 4.5 here is equivalent to Theorem 5.3 of Chapter IV.
Theorem 4.6. IfS and T are subspaces of V, then (S + T) 1 = S^ n T>
and(S Hi) 1 =S L +T 1 .
proof. This theorem is equivalent to Theorem 4.4 of Chapter IV. □
Theorem 4.7. For each conjugate bilinear form f, there is a linear trans
formation a such thatf(oi, /S) = (a, o(0))for all a, e V.
proof. For a fixed a e V, /(a, 0) is linear in /?. Thus by Theorem 3.1
there is a unique V e V such that /(ex, 0) = (rj, 0) for all /J 6 V. Define
ff *( a ) _ ^ a * i s linear since (or*(flia 1 + a 2 <x. z ), 0) =f(a 1 a. 1 + a 2 a. 2 , 0) —
fii/(«i. P) + *tfi"*> P) = ^(°*M> P) + a 2 (a*(* 2 ), 0) = ( ai a*M +
a (T*(a 2 ), j8). Let <r** = a be the linear transformation of which a* is the
adjoint 2 'Then/(a, 0) = (a*(a), 0) = (a, <r(/?)). □
192 Orthogonal and Unitary Transformations, Normal Matrices  V
We shall call a the linear transformation associated with the conjugate
bilinear form /. The eigenvalues and eigenvectors of a conjugate bilinear
form are defined to be the eigenvalues and eigenvectors of the associated
linear transformation. Conversely, to each linear transformation a there
is associated a conjugate bilinear form (a, cr(/?)), and we shall also freely
transfer terminology in the other direction. Thus a linear transformation
will be called symmetric, or skewsymmetric, etc., if it is associated with a
symmetric, or skewsymmetric bilinear form.
Theorem 4.8. The conjugate bilinear form f and the linear transformation a
for which /(a, (S) = (a, o((5)) a™ represented by the same matrix with respect
to an orthonormal basis.
proof. Let X = {£ l5 . . . , i n } be an orthonormal basis and let A = [a iS ]
be the matrix representing a with respect to this basis. Then /(!*, £,) =
(£„ *(£,)) = (£«, 2£=i «mW = 11=k ««(**. £*) = a u . U
A linear transformation is called selfadjoint if cr* = a. Clearly, a linear
transformation is selfadjoint if and only if the matrix representing it (with
respect to an orthonormal basis) is Hermitian. However, by means of
Theorem 4.7 selfadjointness of a linear transformation can be related to the
Hermitian character of a conjugate bilinear form without the intervention
of matrices. Namely, if/ is a Hermitian form then (a* (a), /?) = (a, cr(/?)) =
/(a, /?) =JW*)= (/?, (7(a)) = ((7(a), 0).
Theorem 4.9. If a and r are linear transformations on V such that
(cr(a), j8) = (r(a), /S)/or a// a, /? £ V, f/*e« a = t.
proof. If (cr(a), 0)  (r(a), 0) = ((<r  r)(a), .0) = for all a, 0, then for
each a and /? = (<? — r)(a) we have (cr — r)(a) 2 = 0. Hence, (a — r)(a) =
for all a and c = t. □
Corollary 4.10. If a and r are linear transformations on V such that
(a, ff(/?)) = (a, r(fi))for all a, ft eV, then a = t. □
Theorem 4.9 provides an independent proof that the adjoint operator
cr* is unique. Corollary 4.10 shows that the linear transformation a corre
sponding to the bilinear form /such that/(a, /?) = (a, cr(/S)) is also unique.
Since, in turn, each linear transformation a defines a bilinear form /by the
formula /(a, /?) = (a, o"(/?)), this establishes a onetoone correspondence
between conjugate bilinear forms and linear transformations.
Theorem 4.11. Let V be a unitary vector space. If a and r are linear trans
formations on V such that (c(a), a) = (t(<x), a) for all a £ V, then a = t.
proof. It can be checked that
(cr(a), 0) = {(cr(a + 0), « + 0) _ (<r(a  ft, a  0)
 /(cr(a + //ff), a + #) + /(<r(a  ip), a  /"£)}. (4.3)
4  The Adjoint Transformation 193
It follows from the hypothesis that ((r(oc), j8) = (r(a), 0) for all a, e V.
Hence, by Theorem 4.9, or = t. □
It is curious to note that this theorem can be proved because of the relation
(4.3), which is analogous to formula (12.4) of Chapter IV. But the analogue
of formula (10. 1) in the real case does not yield the same conclusion. In fact,
if V is a vector space over the real numbers and a is skewsymmetric, then
(o(oc), a) = (a, <r*(a)) = (a, <r(a)) = (a, <x(a)) = (or (a), a) for all a.
Thus (o(a), a) = for all a. In the real case the best analogue of this theorem
is that if (cr(a), a) = (r(a), a) for all a e V, then a + a* = r + t*, or a and r
have the same symmetric part.
EXERCISES
1. Show that ((Tr)* = T*or*.
2. Show that if a* a = 0, then a = 0.
3. Let a be a skewsymmetric linear transformation on a vector space over the
real numbers. Show that a* = — a.
4. Let/be a skewHermitian form— that is, /(a, 0) = /(£, a)— and let er be the
associated skewHermitian linear transformation. Show that a* = —a.
5. Show that eigenvalues of a real skewsymmetric linear transformation are
either or pure imaginary. Show that the same is true for a skewHermitian
linear transformation.
6. For what kind of linear transformation a is it true that (I, <r(£)) = for all
f eV?
7. For what kind of linear transformation a is it true that <r(f) £ I 1 for all 1 6 V?
8. Show that if W is an invariant subspace under a, then W 1 is an invariant
subspace under a*.
9. Show that if a is selfadjoint and W is invariant under a, then Wi is also
invariant under a.
10. Let 77 be the projection of V onto S along T. Let tt* be the adjoint of n.
Show that tt* is the projection of V onto T 1  along S x .
11. Let W = a(V). Show that W 1 is the kernel of a*.
12. Show that a and <r* have the same rank.
13. Let W = o(V). Show that <r*(V) = <r*(W).
14. Show that <r*(V) = cr*cr(V). Show that a(V) = oo*(V).
15. Show that if o*o = a*a, then o*(V) = a(V).
16. Show that if a*a = aa*, then a and o* have the same kernel.
17. Show that a + a* is selfadjoint.
18. Show that if o + a* = 0, then <r is skewsymmetric, or skewHermitian.
194 Orthogonal and Unitary Transformations, Normal Matrices  V
19. Show that a  a* is skewsymmetric, or skewHermitian.
20. Show that every linear transformation is the sum of a selfadjoint trans
formation and a skewHermitian transformation.
21. Show that if aa* = a* a, then Im(ff) is an invariant subspace under a. In
fact, show that a n (V) = o(V) for all n > 1.
22. Show that if o is a scalar transformation, that is a(a) = aa, then cr*(a) = aa.
5 I Orthogonal and Unitary Transformations
Definition. A linear transformation of V into itself is called an isometry if
it preserves length; that is, a is an isometry if and only if cr(a) = aj for
all a g V. An isometry in a vector space over the real numbers is called an
orthogonal transformation. An isometry in a vector space over the complex
numbers is called a unitary transformation. We try to save duplication and
repetition by treating the real and complex cases together whenever possible.
Theorem 5.1. A linear transformation a of V into itself is an isometry if
and only if it preserves the inner product; that is, if and only if (a, /?) =
(<r(a), o{P))for all a, fi e V.
proof. Certainly, if a preserves the inner product then it preserves
length since a(a) 2 = (<r(a), <r(a)) = (a, a) = a 2 .
The converse requires the separation of the real and complex cases. For
an inner product over the real numbers we have
(a, ft = K(a + t a + 0) _ (a, a ) _ (fa £)}
= *{a + £ 2  la 2  £ 2 }. (5.1)
For an inner product over the complex numbers we have
(a, ft = i{(a + 0, a + /8)  (a  0, a  0)
 /(a + ifi, a + i/8) + /(a  #, a  ifi)}
= idla + £ 2   a  £ 2  /  a + i/8» + / a  #f}. (5.2)
In either case, any linear transformation which preserves length will preserve
the inner product. □
Theorem 5.2. A linear transformation a of V into itself is an isometry if
and only if it maps an orthonormal basis onto an orthonormal basis.
proof. It follows immediately from Theorem 5.1 that if a is an isometry,
then a maps every orthonormal set onto an orthonormal set and, therefore,
an orthonormal basis onto an orthonormal basis.
On the other hand, let X = {f l9 . . . , f n } be any orthonormal basis which
is mapped by a onto an orthonormal basis {c(^ 1 ), . . . , <r(£ n )}. For an
6  Orthogonal and Unitary Matrices 195
arbitrary vector a £ V, a = 2?=i x i%i> we nave
Ka) 2 = (cr(a),<r(a))
i=l 3=1
n n
i=l 3=1
= i^=H 2 . (5.3)
Thus a preserves length and it is an isometry. □
Theorem 5.3. a is an isometry if and only if a* = (f 1 .
proof. If a is an isometry, then 0(a), cr(#)) = (a, /S) for all a, ft e V. By
the definition of <r*, (a, 0) = (<r*[<r(a)], #) = (<r*<r(a), #)• Since this
equation holds for all /9 e V, c*or(a) is uniquely defined and a*a(jx) = a. Thus
a* a is the identity transformation, that is, a* = cr 1 .
Conversely, suppose that a* = a 1 . Then (<r(a), a(fi)) = (<r*[<r(a)], /S) =
(<r*<r(a), /3) = (a, fi) for all a, e V, and a is an isometry. □
EXERCISES
1. Let a be an isometry and let A be an eigenvalue of a. Show that A = 1.
2. Show that the real eigenvalues of an isometry are ±1.
3. Let X = {f l5  2 } be an orthonormal basis of V. Find an isometry that maps
1
£x onto ^= (fi + f 2 ).
4. Let X = {fi, l 2 » ^3} be an orthonormal basis of V. Find an isometry that
maps fi onto J(f x + 2I 2 + 2^)
6 I Orthogonal and Unitary Matrices
Let a be an isometry and let U = [i/ w ] be a matrix representing a with
respect to an orthonormal basis X = {k t , . . . , IJ. Since a is an isometry,
the set X' = {a(^), . . . , <r(£J} must also be orthonormal. Thus
*=i «=i /
n / n \
n
196 Orthogonal and Unitary Transformations, Normal Matrices  V
This is equivalent to the matrix equation U*U = I, which also follows
from Theorem 5.3.
It is also easily seen that if U*U = I, then a must map an orthonormal
basis onto an orthonormal basis. By Theorem 5.2 a is then an isometry.
Thus,
Theorem 6.1. A matrix U whose elements are complex numbers represents
a unitary transformation {with respect to an orthonormal basis) if and only if
U* = U l . A matrix with this property is called a unitary matrix. □
If the underlying field of scalars is the real numbers instead of the complex
numbers, then U is real and U* = U T . Nothing else is really changed and
we have the corresponding theorem for vector spaces over the real numbers.
Theorem 6.2. A matrix U whose elements are real numbers represents an
orthogonal transformation (with respect to an orthonormal basis) if and only
if JJ T = U~ x . A real matrix with this property is called an orthogonal
matrix. □
As is the case in Theorems 6.1 and 6.2, quite a bit of the discussion of
unitary and orthogonal transformations and matrices is entirely parallel.
To avoid unnecessary duplication we discuss unitary transformations and
matrices and leave the parallel discussion for orthogonal transformations
and matrices implicit. Up to a certain point, an orthogonal matrix can be
considered to be a unitary matrix that happens to have real entries. This
viewpoint is not quite valid because a unitary matrix with real coefficients
represents a unitary transformation, an isometry on a vector space over
the complex numbers. This viewpoint, however, leads to no trouble until
we make use of the algebraic closure of the complex numbers, the property
of complex numbers that every polynomial equation with complex co
efficients possesses at least one complex solution.
It is customary to read equations (6.1) as saying that the columns of U are
orthonormal. Conversely, if the columns of U are orthonormal, then
U* = U" 1 and U is unitary. Also, U* as a left inverse is also a right inverse ;
that is, UU* = /. Thus,
n n
2 "*fc"7fc = d a = 2 uJkUik (62)
fc=l k=l
Thus U is unitary if and only if the rows of U are orthonormal. Hence,
Theorem 6.3. Unitary and orthogonal matrixes are characterized by the
property that their columns are orthonormal. They are equally characterized
by the property that their rows are orthonormal. □
Theorem 6.4. The product of unitary matrices is unitary. The product of
orthogonal matrices is orthogonal.
6  Orthogonal and Unitary Matrices 197
proof. This follows immediately from the observation that unitary and
orthogonal matrices represent isometries, and one isometry followed by
another results in an isometry. □
A proof of Theorem 6.4 based on the characterizing property U* = U~ l
(or U T — JJ 1 for orthogonal matrices) is just as brief. Namely, (1/^)* =
u*u? = uw^ = (u x u 2 )k
Now suppose that X = {f l9 ...,!„} and X' = {£[, . . . , g£ are two
orthonormal bases, and that P = [p tj ] is the matrix of transition from the
basis X to the basis X'. By definition,
i; = JU^. (63)
Thus,
\t=l s=l
n to
= 2 Pa 2 P S fc(^> f «)
«=1 s=l
TO
This means the columns of P are orthonormal and P is unitary (or orthogonal).
Thus we have
Theorem 6.5. The matrix of transition from one orthonormal basis to
another is unitary (or orthogonal if the underlying field is real). □
We have seen that two matrices representing the same linear transformation
with respect to different bases are similar. If the two bases are both ortho
normal, then the matrix of transition is unitary (or orthogonal). In this case
we say that the two matrices are unitary similar (or orthogonal similar).
The matrices A and A' are unitary (orthogonal) similar if and only if there
exists a unitary (orthogonal) matrix P such that A' = P~ X AP = P*AP
(A' = P^AP = P T AP).
If H and H' are matrices representing the same conjugate bilinear form
with respect to different bases, they are Hermitian congruent and there
exists a nonsingular matrix P such that H' = P*HP. P is the matrix of
transition and, if the two bases are orthonormal, P is unitary. Then H' =
P*HP = P~ X HP. Hence, if we restrict our attention to orthonormal bases
in vector spaces over the complex numbers, we see that matrices representing
linear transformations and matrices representing conjugate bilinear forms
transform according to the same rules ; they are unitary similar.
198
Orthogonal and Unitary Transformations, Normal Matrices  V
If B and B' are matrices representing the same real bilinear form with
respect to different bases, they are congruent and there exists a nonsingular
matrix P such that B' = P T BP. P is the matrix of transition and, if the two
bases are orthonormal, P is orthogonal. Then B' = P T BP = P _1 BP.
Hence, if we restrict our attention to orthonormal bases in vector spaces
over the real numbers, we see that matrices representing linear transforma
tions and matrices representing bilinear forms transform according to the
same rules; they are orthogonal similar.
In our earlier discussions of similarity we sought bases with respect to
which the representing matrix had a simple form, usually a diagonal form.
We were not always successful in obtaining a diagonal form. Now we
restrict the set of possible bases even further by demanding that they be
orthonormal. But we can also restrict our attention to the set of matrices
which are unitary (or orthogonal) similar to diagonal matrices. It is fortunate
that this restricted class of matrices includes a rather wide range of cases
occurring in some of the most important applications of matrices. The main
goal of this chapter is to define and characterize the class of matrices unitary
similar to diagonal matrices and to organize computational procedures by
means of which these diagonal matrices and the necessary matrices of
transition can be obtained. We also discuss the special cases so important
in the applications of the theory of matrices.
EXERCISES
1. Test the following matrices for orthogonality. If a matrix is orthogonal,
find its inverse.
1
2
V3
2
(b)
1
2
V3
2
(c)
"0.6
0.8"
V3
1
V3
1
_0.8
0.6
2
2
2
2
2. Which of the following matrices are unitary ?
(a)
1 +1
1 r
2
1 1
2
1 +/
(b)
"1 f
J 1
(c)
"1
J
—1
1_
2
2 _
3. Find an orthogonal matrix with (1/^2, 1/^2) in the first column. Find an
orthogonal matrix with (£, f , §) in the first column.
4. Find a symmetric orthogonal matrix with (J, , §) in the first column. Com
pute its square.
7  Superdiagonal Form
199
5. The following matrices are all orthogonal. Describe the geometric effects
in real Euclidean 3space of the linear transformations they represent.
(«)
id)
1
0"
(b)
~l
0"
(c)
"1
0"
1
1
1
1
1
1
"cos —sin 0"
sin cos
(e)
cos —sin 0"
sin cos
1J _ 01
Show that these five matrices, together with the identity matrix, each have different
eigenvalues (provided is not 0° or 180°), and that the eigenvalues of any third
order orthogonal matrix must be one of these six cases.
6. If a matrix represents a rotation of R 2 around the origin through an angle of
0, then it has the form
"cos —sin 0"
sin cos
A(8) =
Show that A{d) is orthogonal. Knowing that A{0) ■ A{ip) = A{B + y>), prove that
sin (0 + y) = sin cos y> + cos sin y>. Show that if U is an orthogonal 2x2
matrix, then U X A{B)U = A(±6).
7. Find the matrix B representing the real quadratic form q{x, y) = ax 2 +
2bxy + cy 2 . Show that the discriminant D = ac — b 2 is the determinant of B.
Show that the discriminant is invariant under orthogonal coordinate changes,
that is, changes of coordinates for which the matrix of transition is orthogonal.
7 I Superdiagonal Form
In this section we restrict our attention to vector spaces (and to matrices)
over the field of complex numbers. We have already observed that not
every matrix is similar to a diagonal matrix. Thus, it is also true that not
every matrix is unitary similar to a diagonal matrix. We later restrict our
attention to a class of matrices which are unitary similar to diagonal matrices.
As an intermediate step we obtain a relatively simple form to which every
matrix can be reduced by unitary similar transformations.
Theorem 2.1. Let a be any linear transformation of V, a finite dimensional
vector space over the complex numbers, into itself There exists an ortho
normal basis of V with respect to which the matrix representing a is in super
diagonal form; that is, every element below the main diagonal is zero.
proof. The proof is by induction on n, the dimension of V. The theorem
says there is an orthonormal basis Y = {rj 1 , . . . , r} n } such that a(r] k ) =
S=i fl ifc*7*' tne important property being that the summation ends with the
fcth term. The theorem is certainly true for n = 1 .
200 Orthogonal and Unitary Transformations, Normal Matrices  V
Assume the theorem is true for vector spaces of dimensions <«. Since
V is a vector space over the complex numbers, a has at least one eigenvalue.
Let X x be an eigenvalue for a and let £[ 5^ 0, \\^\\ = 1, be a corresponding
eigenvector. There exists a basis, and hence an orthonormal basis, with
£J as the first element. Let the basis be X' = {£{, ...,£,} and let W be the
subspace spanned by {£' 2 , . . . , ^}. Wis the subspace consisting of all vectors
orthogonal to g[. For each a = 2*=i a^' define r(a) = 2? = 2 fl <£< e W 
Then to 1 restricted to W is a linear transformation of W into itself. According
to the induction assumption, there is an orthonormal basis {rj 2 , . . . , rj n }
of W such that for each rj k , ra(r) k ) is expressible in terms of {rj 2 , . . . , t] k }
alone. We see from the way r is defined that a(r] k ) is expressible in terms of
{!{, r) 2 , . . . , rj k } alone. Let r\ x = l' v Then V = {rj lt rj 2 , . . . , ^J is the re
quired basis.
alternate proof. The proof just given was designed to avoid use of the
concept of adjoint introduced in Section 4. Using that concept, a very much
simpler proof can be given. This proof also proceeds by induction on n. The
assertion for n = 1 is established in the same way as in the first proof given.
Assume the theorem is true for vector spaces of dimension <«. Since V is a
vector space over the complex numbers, a* has at least one eigenvalue. Let X n
be an eigenvalue for a* and let rj n , \\rj n \\ = 1, be a corresponding eigenvector.
Then by Theorem 4.4, W = (rjn) 1  is an invariant subspace under a. Since
rj n 7* 0, W is of dimension n — 1. According to the induction assumption,
there is an orthonormal basis {rj x , . . . , ^ n _i} of W such that o(r) k ) =
2=i ^kVi* for fc = 1, 2, . . . , n  1. However, {^, . . . , rj n } is also an
orthonormal basis of U and o(rj n ) = ^ =1 a^v}^ for k = 1 , . . . , n. □
Corollary 7.2. Over the field of complex numbers, every matrix is unitary
similar to a superdiagonal matrix. □
Theorem 7.1 and Corollary 7.2 depend critically on the assumption that
the field of scalars is the field of complex numbers. The essential feature
of this condition is that it guarantees the existence of eigenvalues and eigen
vectors. If the field of scalars is not algebraically closed, the theorem is
simply not true.
Corollary 7.3. The diagonal terms of the superdiagonal matrix representing
a are the eigenvalues of a.
proof. If A = [a u ] is in superdiagonal form, then the characteristic
polynomial is (a n — x)(a 22 — x) • •  (a nn — x). a
EXERCISES
1. Let a be a linear transformation mapping U into V. Let A be any basis of U
whatever. Show that there is an orthonormal basis 8 of V such that the matrix
8  Normal Matrices 201
representing a with respect to A and 8 is in superdiagonal form. (In this case
where U and V need not be of the same dimension so that the matrix representing
a need not be square, by superdiagonal form we mean that all elements below the
main diagonal are zeros.)
2. Let a be a linear transformation on V and let V = {^ l5 . . . , t] n } be an ortho
normal basis such that the matrix representing a with respect to V is in super
diagonal form. Show that the matrix representing a* with respect to V is in sub
diagonal form; that is, all elements above the main diagonal are zeros.
3. Let or be a linear transformation on V. Show that there is an orthonormal
basis Y of V such that the matrix representing a with respect to Y is in subdiagonal
form.
8 I Normal Matrices
It is possible to give a necessary and sufficient condition that a matrix be
unitary similar to a diagonal matrix. The real value in establishing this
condition is that several important types of matrices do satisfy this condition.
Theorem 8.1. A matrix A in superdiagonal form is a diagonal matrix if
and only if A* A = AA*.
proof. Let A = [a w ] where a ti = if / > j. Suppose that A* A = AA*.
This means, in particular, that
n n
X fi « fl « = lMii ( 8 l)
But since a i} = for / > j, this reduces to
i n
l\a H \ 2 =2\a ik \ 2 . (8.2)
j'=l k=i
Now, if A were not a diagonal matrix, there would be a first index / for
which there exists an index k > i such that a ik ¥" 0. For this choice of the
index / the sum on the left in (8.2) reduces to one term while the sum on the
right contains at least two nonzero terms. Thus,
t\a H \ 2 =\a H \ 2 = l\a ik \\ (8.3)
which is a contradiction. Thus A must be a diagonal matrix.
Conversely, if A is a diagonal matrix, then clearly A* A = AA*. n
A matrix A for which A* A = AA * is called a normal matrix.
Theorem 8.2. A matrix is unitary similar to a diagonal matrix if and only
if it is normal.
202
Orthogonal and Unitary Transformations, Normal Matrices  V
proof. If A is a normal matrix, then any matrix unitary similar to A is
also normal. Namely, if U is unitary, then
(U*AU)*(U*AU) = U*A*UU*AU
= U*A*AU
= U*AA*U
= U*AUU*A*U.
= (U*AU)(U*AU)*. (8.4)
Thus, if A is normal, the superdiagonal form to which it is unitary similar
is also normal and, hence, diagonal. Conversely, if A is unitary similar to
a diagonal matrix, it is unitary similar to a normal matrix and it is therefore
normal itself. □
Theorem 8.3. Unitary matrices and Hermitian matrices are normal.
proof. If U is unitary then U*U = U~ X U = UU 1 = UU*. If H is
Hermitian then H*H = HH = HH*. □
EXERCISES
1. Determine which of the following matrices are orthogonal, unitary, symmetric
Hermitian, skewsymmetric, skewHermitian, or normal.
(a)
(d)
(g)
0')
"1 
2"
(b)
"1
i
(c) f
"1
f
2 1
i
1
i
2_
"o r
(e)
"0
r
(/)
"
1 1  f
1 o_
1
1
+ / 3
"1 2
2"
(h)
"1 2
2"
(0
i 2 2
1
i
3
2 2
1
2 1
2_
2 1
2_
"i r
(*)
"0
l r
1
1
1
1
i
1
3
2"
3
2. Which of the matrices of Exercise 1 are unitary similar to diagonal matrices?
3. Show that a real skewsymmetric matrix is normal.
4. Show that a skewHermitian matrix is normal.
5. Show by example that there is a skewsymmetric complex matrix which is
not normal.
6. Show by example that there is a symmetric complex matrix which is not
normal.
7. Find an example of a normal matrix which is not Hermitian or unitary.
8. Show that if M = A + Bi where A and B are real and symmetric, then M
is normal if and only if A and B commute.
9  Normal Linear Transformations 203
9 I Normal Linear Transformations
Theorem 9.1. If there exists an orthonormal basis consisting of eigen
vectors of a linear transformation a, then a* a = aa*.
proof. Let X = {£ l5 . . . , £J be an orthonormal basis consisting of
eigenvectors of a. Let X t be the eigenvalue corresponding to ^. Then.
(**(£,), *,) = (*„ *(£,)) = (*„ A,,) = V«y = ^» = *<(£<» W = <¥<' **)•
For a fixed < this equation holds for all  3 and, hence, (or*(£j), a) = (!<£„ a)
for all a e V. This means <r*(f <) = !<£ < and £< is an eigenvector of <r* with
eigenvalue I,. Then <ra*(f < ) = trfof*) = I<A,f, = <r*tf(£<) Since aa* =
a*a on a basis of V, aa* = a*a on all of V. □
A linear transformation a for which a* a = aa* is called a normal linear
transformation. Clearly, a linear transformation is normal if and only if the
matrix representing it (with respect to an orthonormal basis) is normal.
In the proof of Theorem 9.1 the critical step is showing that an eigenvector
of a is also an eigenvector of a*. The converse is also true.
Theorem 9.2. If  is an eigenvector of a normal linear transformation a
corresponding to the eigenvalue A, then £ is an eigenvector of a* corresponding
to I.
proof. Since a is normal (cr(£), cr(£)) = (<t*<t(£)> f) = ( ffff *(f)> f) =
(ff*(£), <**(£))> Since £ is an eigenvector of a corresponding to A, <r(£) = A£
so that
= <r(£)  A£ 2 = (<r(£)  A*, <;(£)  A!)
= (<r(£), <r(£))  A(£, <r(£))  A(or(D, £) + lA(, £)
= (<r*(£), **(0)  A(or*(), £)  A(£, a*(£)) + AA(, £)
= (<r*(£)  A£, <r*(£)  A£)
= cr*(£)A£ 2 . (9.1)
Thus o*(£)  A£ = 0, or <r*(£) = A£. n
Theorem 9.3. For a normal linear transformation, eigenvectors corre
sponding to different eigenvalues are orthogonal.
proof. Suppose a{i x ) = A^ and a(£ 2 ) = ^£2 where A x 5^ A 2 . Then
A 2 (li, £ 2 ) = (£1, A 2 £ 2 ) = (£ l5 <r(£ 2 )) = (*•(£,), £ 2 ) = (A^, £ 2 ) = A^, £ 2 ).
Thus (A x  A 2 )(£ l5 £ 2 ) = 0. Since l x  A 2 5* we see that (£ l5 £ 2 ) = 0;
that is, £ x and £ 2 are orthogonal. □
Theorem 9.4. If a is normal, then (a(a), a(j8)) = (a*(a), a*(]8)) /or all
a, e V.
PROOF. (cr(a), <r(0)) = (a*a(a), 8) = (orcr*(a), 0) = (a*(a), <r*(0)). D
Corollary 9.5. If a is normal, \\a(a.)\\ = <r*(a)) /or a// a e V. O
204 Orthogonal and Unitary Transformations, Normal Matrices  V
Theorem 9.6. If ((7(a), a(fi)) = ((7*(a), <r*(/3)) for all aJeV, then a is
normal.
proof, (a, ac7*03)) = (ff*(a), (7*(/3)) = (or(a), <r(0)) = (a, o*o(P)) for all
aJeV. By Corollary 4.10, (7(7* = or*(T and (7 is normal. □
Theorem 9.7. If \\a(<x)\\ = <r*(a) /or a// a g V, then a is normal.
proof. We must divide this proof into two cases :
1. V is a vector space over F, a subfield of the real numbers. Then
((7(a), <r(fl) = i{ll*(« + £)U 2  lk(a  /8)»}.
It then follows from the hypothesis that (<r(a), <x(j0)) = (<r*(a), <r*(0)) for all
a, /3 G V, and c is normal.
2. V is a vector space over F, a nonreal normal subfield of the complex
numbers. Let a e F be chosen so that a ^ a. Then
((7(a), (7(/S)) = *
2(a — a)
X {a (7(a + 0) 2  a Ka  £) 2  (7(a + a£) 2 + (7(a  a£) 2 }.
Again, it follows that a is normal. □
Theorem 9.8. If a is normal then K(o) = K(o*).
proof. Since <y(a) = <r*(a), (7(a) = if and only if (7*(a) = 0. D
Theorem 9.9. If a is normal, K{a) = lm(o) L .
proof. By Theorem 4.5 K(o*) = Im((7) x , and by Theorem 9.8 K(a) —
K(o*). D
Theorem 9.10. If a is normal, Im a = Im a*.
proof. Im (7 = ^(a) 1 = Im a*. O
Theorem 9.11. If a is a normal linear transformation and W is a set of
eigenvectors of a, then W L is an invariant subspace under a.
proof, a g W 1 if and only if (£, a) = for all £ e W. But then (, cr(a)) =
(<r*(), a) = (A, a) = A(£, a) = 0. Hence, (7(a) e W x and W 1 is invariant
under cr. D
Notice it is not necessary that W be a subspace, it is not necessary that
W contain all the eigenvectors corresponding to any particular eigenvalue,
and it is not necessary that the eigenvectors in W correspond to the same
eigenvalue. In particular, if I is an eigenvector of a, then {£}■> is an invariant
subspace under a.
Theorem 9.12. Let V be a vector space with an inner product, and let a be a
normal linear transformation of V into itself. If W is a subspace which is
invariant under both a and g*, then a is normal on W.
9  Normal Linear Transformations 205
proof. Let a denote the linear transformation of W into itself induced
by a. Let o* denote the adjoint of a on W. Then for all, a, /? e W we have
(or*(a), 0) = (a, a0)) = (a, cr(0)) = (or* (a), /?).
Since ((a* — cr*)(a), /3) = for all a, /S £ W, a* and or* coincide on W.
Thus ff*cr = a*a = era* = oq* on W, and <r is normal. □
Theorem 9.13. Let V be a finite dimensional vector space over the complex
numbers, and let a be a normal linear transformation on V. If W is invariant
under a, then W is invariant under a* and a is normal on W.
proof. By Theorem 4.4, W 1  is invariant under a*. Let a* be the
restriction of a* with W 1 as domain and codomain. Since W is also a finite
dimensional vector space over the complex numbers, a* has at least one
eigenvalue A and corresponding to it a nonzero eigenvector £. Thus o"*() =
A! = o*(). Thus, we have found an eigenvector for a* in W 1 .
Now proceed by induction. The theorem is certainly true for spaces of
dimension 1. Assume the theorem holds for vector spaces of dimension <«.
By Theorem 9.2, £ is an eigenvector of a. By Theorem 9.1 1 , (I) 1 is invariant
under both a and a*. By Theorem 9.12, a is normal on (I) 1 . Since (!) <=■
WL, W c= (I) 1 . Since dim (I) 1 = n — 1 , the induction assumption applies.
Hence, a is normal on W and W is invariant under a*. □
Theorem 9.13 is also true for a vector space over any subfield of the complex
numbers, but the proof is not particularly instructive and this more general
form of Theorem 9.13 will not be needed later.
We should like to obtain a converse of Theorem 9.1 and show that a normal
linear transformation has enough eigenvectors to make up an orthonormal
basis. Such a theorem requires some condition to guarantee the existence
of eigenvalues or eigenvectors. One of the most important general conditions
is to assume we are dealing with vector spaces over the complex numbers.
Theorem 9.14. If V is a finite dimensional vector space over the complex
numbers and a is a normal linear transformation, then V has an orthonormal
basis consisting of eigenvectors of a.
proof. Let n be the dimension of V. The theorem is certainly true for
n = 1, for if {fi} is a basis a{^ = a^ x .
Assume the theorem holds for vector spaces of dimension <«. Since V
is a finite dimensional vector space over the complex numbers, a has at
least one eigenvalue A l5 and corresponding to it a nonzero eigenvector £ x
which we can take to be normalized. By Theorem 9.1 1 , {li} 1 is an invariant
subspace under a. This means that a acts like a linear transformation on
{£i} ± when we confine our attention to {^i} 1 . But then a is also normal on
206 Orthogonal and Unitary Transformations, Normal Matrices  V
{it} 1 . Since l^} 1  is of dimension n — 1, our induction assumption applies
and {fx} x has an orthonormal basis {£ 2 > • • • , f J consisting of eigenvectors
of a. {£j,  a , . . . , !J is the required orthonormal basis of V consisting of
eigenvectors of a. □
We can observe from examining the proof of Theorem 9.1 that the con
clusion that a and o* commute followed immediately after we showed that
the eigenvectors of a were also eigenvectors of cr*. Thus the following
theorem follows immediately.
Theorem 9.15. If there exists a basis (orthonormal or not) consisting of
vectors which are eigenvectors for both a and r, then ot = to. □
Any possible converse to Theorem 9.15 requires some condition to ensure
the existence of the necessary eigenvectors. In the following theorem we
accomplish this by assuming that the field of scalars is the field of complex
numbers, any set of conditions that would imply the existence of the eigen
vectors could be substituted.
Theorem 9.16. Let V be a finite dimensional vector space over the complex
numbers and let a and r be normal linear transformations on V. If or = to,
then there exists an orthonormal basis consisting of vectors which are eigen
vectors for both o and r.
proof. Suppose or = to. Let A be an eigenvalue of o and let S(A) be
the eigenspace of o consisting of all eigenvectors of o corresponding to A.
Then for each I e S(A) we have crr(£) = to(£) = t(A£) = At(£). Hence,
t(£) e S(A). This shows that S(A) is an invariant subspace under t; that is,
t confined to S(A) can be considered to be a normal linear transformation of
S(X) into itself. By Theorem 9.14 there is an orthonormal basis of S(A) con
sisting of eigenvectors of r. Being in S(A) they are also eigenvectors of a.
By Theorem 9.3 the basis vectors obtained in this way in eigenspaces
corresponding to different eigenvalues of o are orthogonal. Again, by
Theorem 9.14 there is a basis of V consisting of eigenvectors of o. This
implies that the eigenspaces of o span V and, hence, the entire orthonormal
set obtained in this fashion is an orthonormal basis of V. □
As we have seen, selfadjoint linear transformations and isometries are
particular cases of normal linear transformations. They can also be char
acterized by the nature of their eigenvalues.
Theorem 9.17. Let V be a finite dimensional vector space over the complex
numbers. A normal linear transformation o on V is selfadjoint if and only
if all its eigenvalues are real.
9  Normal Linear Transformations 207
proof. Suppose a is selfadjoint. Let A be an eigenvalue for a and let 
be an eigenvector corresponding to A. Then cr(£) 2 = (<?(£), °"(£)) =
(a*(£), ff(£)) = A 2  HI 2 . Thus A 2 is real nonnegative and A is real.
On the other hand, suppose a is a normal linear transformation and that
all its eigenvalues are real. Since a is normal there exists a basis X =
{£ l5 . . . , £ n } of eigenvectors of <r. Let A i be the eigenvalue corresponding
to j. Then c*( J ) = A^ = A^ = cr(£,). Since <7* coincides with a on
a basis of V, a = o  * on all of V. □
Theorem 9.18. Let V be a finite dimensional vector space over the complex
numbers. A normal linear transformation a on V is an isometry if and only
if all its eigenvalues are of absolute value 1.
proof. Suppose a is an isometry. Let A be an eigenvalue of a and let £
be an eigenvector corresponding to A. Then £ 2 = or() 2 = (<r(£),
<r(D) = (A£, A) = A 2 (, ). Hence A 2 = 1.
On the other hand suppose a is a normal linear transformation and that
all its eigenvalues are of absolute value 1 . Since a is normal there exists a
basis X = {£ 1} ...,£„} of eigenvectors of c. Let A t be the eigenvalue
corresponding to f<. Then (cr^), (7(1,)) = (A^, A,£ 3 ) = A^C^, ,) = <5 i; .
Hence, a maps an orthonormal basis onto an orthonormal basis and it is
an isometry. □
EXERCISES
1. Prove Theorem 9.2 directly from Corollary 9.5.
2. Show that if there exists an orthonormal basis such that a and t are both
represented by diagonal matrices, then ot = to.
3. Show that if a and r are normal linear transformations such that ar = to,
then there is an orthonormal basis of V such that the matrices representing a and
t are both diagonal; that is, a and t can be diagonalized simultaneously.
4. Show that the linear transformation associated with a Hermitian form is
selfadjoint.
5. Let /be a Hermitian form and let a be the associated linear transformation.
Let X = {£ x , . . . , £„} be a basis of eigenvectors of a (show that such a basis
exists) and let {A l5 . . . , A n } be the corresponding eigenvalues. Let a = 2?=i a i£*
and =2? =1 bi£i be arbitrary vectors in V. Show that /(a, #) = ^7=1 <*AV
6. (Continuation) Let ^ be the Hermitian quadratic form associated with the
Hermitian form/. Let S be the set of all unit vectors in V; that is, a e S if and only
if  a  =1. Show that the maximum value of ^(a) for a G S is the maximum eigen
value, and the minimum value of ^(a) for a e S is the minimum eigenvalue. Show
that q(<x) # for all nonzero a E V if all the eigenvalues of/ are nonzero and of
the same sign.
208 Orthogonal and Unitary Transformations, Normal Matrices  V
7. Let ff be a normal linear transformation and let {A x , . . . , A fc } be the distinct
eigenvalues of a. Let M* be the subspace of eigenvectors of a corresponding to X t .
Show that V = M x _L ■■•_]_ M*.
8. (Continuation) Let ^ be the projection of V onto M* along M^. Show that
1 = tt x + ■ ■ • + TT k . Show that a = X 1 tt 1 + ■ ■ ■ + X k ir k . Show that a r = V*a +
• • • + X k r iz k . Show that if p{x) is a polynomial, then/?(<r) = 2* =1 p{K)^i
10 I Hermitian and Unitary Matrices
Although all the results we state in this section have already been obtained,
they are sufficiently useful to deserve being summarized separately. In this
section we are considering matrices whose entries are complex numbers.
Theorem 10.1. If H is Hermitian, then
(1) H is unitary similar to a diagonal matrix D.
(2) The elements along the main diagonal of D are the eigenvalues of H.
(3) The eigenvalues of H are real.
Conversely, ifH is normal and all its eigenvalues are real, then H is Hermitian.
proof. We have already observed that a Hermitian matrix is normal
so that (1) and (2) follow immediately. Since D is diagonal and Hermitian,
D = D* = D and the eigenvalues are real.
Conversely, if H is a normal matrix with real eigenvalues, then the diagonal
form to which it is unitary similar must be real and hence Hermitian. Thus
H itself must be Hermitian. □
Theorem 10.2. If A is unitary, then
(1) A is unitary similar to a diagonal matrix D.
(2) The elements along the main diagonal of D are the eigenvalues of A.
(3) The eigenvalues of A are of absolute value 1.
Conversely, if A is normal and all its eigenvalues are of absolute value 1 ,
then A is unitary.
proof. We have already observed that a unitary matrix is normal so that
(1) and (2) follow immediately. Since D is also unitary, DD = D*D = I
so that AJ 2 = X t Xi = 1 for each eigenvalue X t .
Conversely, if A is a normal matrix with eigenvalues of absolute value 1,
then from the diagonal form D we have D*D = DD = / so that D and A
are unitary. □
Corollary 10.3. If A is orthogonal, then
(1) A is unitary similar to a diagonal matrix D.
(2) The elements along the main diagonal of D are the eigenvalues of A.
(3) The eigenvalues of A are of absolute value 1. □
1 1 I Real Vector Spaces
209
This is a conventional statement of this corollary and in this form it is
somewhat misleading. If A is a unitary matrix that happens to be real,
then this corollary says nothing that is not contained in Theorem 10.2.
A little more information about A and its eigenvalues is readily available.
For example, the characteristic equation is real so that the eigenvalues occur
in conjugate pairs. An orthogonal matrix of odd order has at least one
real eigenvalue, etc. If A is really an orthogonal matrix, representing an
isometry in a vector space over the real numbers, then the unitary matrix
mentioned in the corollary does not necessarily represent a permissible
change of basis. An orthogonal matrix is not always orthogonal similar
to a diagonal matrix. As an example, consider the matrix representing a
90° rotation in the Euclidean plane. However, properly interpreted, the
corollary is useful.
EXERCISES
1. Find the diagonal matrices to which the following matrices are unitary
similar. Classify each as to whether it is Hermitian, unitary, or orthogonal.
(«)
(fO
1 +1
1 /
(b)
"1
J
—f
1_
(c)
2
1 i
2
1 +/
2 J
2
0.6 0.8"
0.8 0.6
" 3
1 i
1 +f
2
(e)
i
2. Let A be an arbitrary square complex matrix. Since A *A is Hermitian, there is
a unitary matrix P such that P*A*AP is a diagonal matrix D. Let F = P*AP.
Show that F*F = D. Show that D is real and the elements of D are nonnegative.
3. Show that every complex matrix can be written as the sum of a real matrix,
and an imaginary matrix; that is, if M is complex, then M = A + Bi where A and
B are real. Show that M is Hermitian if and only if A is symmetric and B is skew
symmetric. Show that M is skewHermitian if and only if A is skewsymmetric and
B is symmetric.
11 I Real Vector Spaces
We now wish to consider linear transformation and matrices in vector
spaces over the real numbers. Much of what has been done for complex
vector spaces can be carried over to real vectors spaces without any difficulty.
We must be careful, however, when it comes to theorems depending on the
210 Orthogonal and Unitary Transformations, Normal Matrices  V
existence of eigenvalues and eigenvectors. In particular, Theorems 7.1 and
7.2 do not carry over as stated. Those parts of Section 8 and 9 which depend
on these theorems must be reexamined carefully before their implications for
real vector spaces can be established.
An examination of the proof of Theorem 7.1 will reveal that the only use
made of any special properties of the complex numbers not shared by the
real numbers was at the point where it was asserted that each linear trans
formation has at least one eigenvalue. In stating a corresponding theorem
for real vector spaces we have to add an assumption concerning the existence
of eigenvalues. Thus we have the following modification of Theorem 7.1
for real vector spaces.
Theorem 11.1 Let V be a finite dimensional vector space over the real
numbers, and let a be a linear transformation on V whose characteristic poly
nomial factors into real linear factors. Then there exists an orthonormal basis
of V with respect to which the matrix representing a is in super diagonal form.
proof. Let n be the dimension of V. The theorem is certainly true for
n = 1.
Assume the theorem is true for real vector spaces of dimensions <«.
Let X x be an eigenvalue for a and let £{ ^ 0, { = 1, be a corresponding
eigenvector. There exists an orthonormal basis with £{ as the first element.
Let the basis be X' = {£[, . . , , £^} and let W be the subspace spanned by
{!;, . . . , f ;}. For each a = j% =1 a^ define r(a) = %U a & G W  Then
to restricted to W is a linear transformation of W into itself.
In the proof of Theorem 7.1 we could apply the induction hypothesis to
tg without any difficulty since the assumptions of Theorem 7.1 applied to
all linear transformations on V. Now we are dealing with a set of linear
transformations, however, whose characteristic polynomials factor into
real linear factors. Thus we must show that the characteristic polynomial
for to factors into real linear factors.
First, consider to as defined on all of V. Since to(£[) = t(A 1 j) = 0,
Tff(a) = Tcr[r(a)] = tot(ol) for all a e V. This implies that (W) fc (a) = rcr fc (a)
since any t to the right of a o can be omitted if there is a t to the left of that a.
Let f(x) be the characteristic polynomial for a. It follows from the
observations of the previous paragraph that Tf{To) = Tf(o) = on V.
But on W, t acts like the identity transformation, so that /(to  ) = when
restricted to W. Hence, the minimum polynomial for to on W divides /(#).
By assumption, f(x) factors into real linear factors so that the minimum
polynomial for to on W must also factor into real linear factors. This
means that the hypotheses of the theorem are satisfied for to on W. By
induction, there is an orthogonal basis {r} 2 , . . . , r} n } of W such that for
each rj k , To{r) k ) is expressible in terms of {tj 2 , . . . , rj k } alone. We see from
11 I Real Vector Spaces 211
the way t is defined that a(rj k ) is' expressible in terms of {^, rj 2 , . . . , rj k }
alone. Let rj t = f[. Then Y = {%, r} 2 , . . . , r] n } is the required basis. □
Since any n X n matrix with real entries represents some linear trans
formation with respect to any orthonormal basis, we have
Theorem 11.2. Let Abe a real matrix with real characteristic values. Then
A is orthogonal similar to a superdiagonal matrix. □
Now let us examine the extent to which Sections 8 and 9 apply to real
vector spaces. Theorem 8.1 applies to matrices with coefficients in any
subfield of the complex numbers and we can use it for real matrices without
reservation. Theorem 8.2 does not hold for real matrices, however. To
obtain the corresponding theorem over the real numbers we must add the
assumption that the characteristic values are real. A normal matrix with real
characteristic values is Hermitian and, being real, it must then be symmetric.
On the other hand a real symmetric matrix has all real characteristic values.
Hence, we have
Theorem 11.3. A real matrix is orthogonal similar to a diagonal matrix if
and only if it is symmetric. □
Because of the importance of real quadratic forms, in many applications
this is a very useful and important theorem, one of the most important of
this chapter. We describe some of the applications in Chapter VI and show
how this theorem is used.
Of the theorems in Section 9 only Theorems 9.14 and 9.16 fail to hold as
stated for real vector spaces. As before, adding the assumption that all the
characteristic values of the linear transformation a are real to the condition
that a is normal amounts to assuming that a is selfadjoint. Hence, the
theorems corresponding to Theorems 9.14 and 9.16 are
Theorem 11.4. If V is a finite dimensional vector space over the real
numbers and a is a selfadjoint linear transformation on V, then V has an
orthonormal basis consisting of eigenvectors of a. □
Theorem 11.5. Let V be a finite dimensional vector space over the real
numbers and let a and r be selfadjoint linear transformations on V. Ifctr = to,
then there exists an orthonormal basis of V consisting of vectors which are
eigenvectors for both a and t. □
Theorem 9.18 must be modified by substituting the words "characteristic
values" for "eigenvalues." Thus,
Theorem 11.6. A normal linear transformation a defined on a real vector
space V is an isometry if and only if all its characteristic values are of absolute
value 1. □
212
Orthogonal and Unitary Transformations, Normal Matrices  V
EXERCISES
1. For those of the following matrices which are orthogonal similar to diagonal
matrices, find the diagonal form.
(a)
{d)
"13
6"
(6)
'1 T
(c)
"i r
6
3_
_5 4_
_/ i_
" 7
4
4"
w
" 13 4 2"
4
8
1
4 13 2
_4
1
8_
2 2 10
" 3
4
2"
Of)
"_4 4 _2 _
4
1
6
44 2
2
6
2_
_2 2 1_
" 1
1
0"
(0
"0 1 2 "
1
1
1 ~l 1
1
_2 1 — V_
"1
2
2"
(*)
~5 2 2"
2
1
2
2 24
_2 
2
1
_2 4
2_.
(/)
(A)
0)
2. Which of the matrices of Exercise 1, Section 8, are orthogonal similar to
diagonal matrices ?
3. Let A and B be real symmetric matrices with A positive definite. There is a
nonsingular matrix P such that P T AP = I. Show that P T BP is symmetric. Show
that there exists a nonsingular matrix Q such that Q T AQ = I and Q T BQ is a
diagonal matrix.
4. Show that every real skewsymmetric matrix A has the form A = P T BP
where P is orthogonal and B 2 is diagonal.
5. Show that if A and B are real symmetric matrices, and A is positive definite,
then the roots of det (B — xA) = are all real.
6. Show that a real skewsymmetric matrix of positive rank is not orthogonal
similar to a diagonal matrix.
7. Show that if A is a real 2x2 normal matrix with at least one element equal
to zero, then it is symmetric or skewsymmetric.
8. Show that if A is a real 2x2 normal matrix with no zero element, then A
is symmetric or a scalar multiple of an orthogonal matrix.
9. Let ffbea skewsymmetric linear transformation on the vector space V over
the real numbers. The matrix A representing a with respect to an orthonormal
12  The Computational Processes 213
basis is skewsymmetric. Show that the real characteristic values of A are zeros. The
characteristic equation may have complex solutions. Show that all complex
solutions are pure imaginary. Why are these solutions not eigenvalues of al
10. (Continuation) Show that a 1 is symmetric. Show that the characteristic
values of A 2 are real. Show that the nonzero eigenvalues of A 2 are negative. Let
— ft 2 be a nonzero eigenvalue of a 2 and let I be a corresponding eigenvector. Define
1
tj to be <r(f). Show that o(rj) = — fig. Show that g and r\ are orthogonal. Show
fi
that r] is also an eigenvector of a 2 corresponding to — fi 2 .
11. (Continuation) Let a be the skewsymmetric linear transformation con
sidered in Exercises 9 and 10. Show that there exists an orthonormal basis of V
such that the matrix representing a has all zero elements except for a sequence of
2x2 matrices down the main diagonal of the form
' —ft k ~
ji k _
where the numbers fi k are defined as in Exercise 10.
12. Let a be an orthogonal linear transformation on a vector space V over the
real numbers. Show that the real characteristic values of or are ±1. Show that any
eigenvector of a corresponding to a real eigenvalue is also an eigenvector of a*
corresponding to the same eigenvalue. Show that these eigenvectors are also
eigenvectors of a + a* corresponding to the eigenvalues ±2.
13. (Continuation) Show that a + a* is selfadjoint. Show that there exists a
basis of eigenvectors of a + a*. Show that if an eigenvector of a + a* is also an
eigenvector of a, then the corresponding eigenvalue is ±2. Let 2fi be an eigenvalue
of or + a* for which the corresponding eigenvector g is not an eigenvector of o.
Show that ft is real and that \/i\ < 1. Show that (g, a(g)) = /<(g, g).
o(g)  /«£
14. (Continuation) Define r\ to be — Show that I and r\ are orthogonal.
VI — ft 2
Show that o(f) = fig + v 7 ! _ ^rj, and a(rj) =  v'l  ffig + fl71 .
15. (Continuation) Let a be the orthogonal linear transformation considered
in Exercises 12, 13, 14. Show that there exists an orthonormal basis of V such that
the matrix representing a has all zero elements except for a sequence of ± l's and/or
2x2 matrices down the main diagonal of the form
"cos 6 k —sin
sin d k cos 6 k \ ,
where fi k = cos 6 k are defined as in Exercise 13.
12 I The Computational Processes
We now summarize a complete set of computational steps which will
effectively determine a unitary (or orthogonal) matrix of transition for
214 Orthogonal and Unitary Transformations, Normal Matrices  V
diagonalizing a given normal matrix. Let A be a given normal matrix.
1. Determine the characteristic matrix C(x) = A — xl.
2. Compute the characteristic polynomial f(x) = det (A — xl).
3. Determine all eigenvalues of A by finding all the solutions of the
characteristic equation f(x) = 0. In any but very special or contrived
examples this step is tedious and lengthy. In an arbitrarily given example
we can find at best only approximate solutions. In that case all the following
steps are also approximate. In some applications special information deriv
able from the peculiarities of the application will give information about the
eigenvalues or the eigenvectors without our having to solve the characteristic
equation.
4. For each eigenvalue X { find the corresponding eigenvectors by solving
the homogeneous linear equations
C(l i )X=0. (12.1)
Each such system of linear equations is of rank less than n. Thus the technique
of Chapter II7 is the recommended method.
5. Find an orthonormal basis consisting of eigenvectors of A. If the
eigenvalues are distinct, Theorem 9.3 assures us that they are mutually
orthogonal. Thus all that must be done is to normalize each vector and the
required orthonormal basis is obtained immediately.
Even where a multiple eigenvalue X t occurs, Theorem 8.2 or Theorem 9.14
assures us that an orthonormal basis of eigenvectors exists. Thus, the
nullity of C(X t ) must be equal to the algebraic multiplicity of k t . Hence,
there is no difficulty in obtaining a basis of eigenvectors. The problem is
that the different eigenvectors corresponding to the multiple eigenvalue X t
are not automatically orthogonal; however, that is easily remedied. All
we need to do is to take a basis of eigenvectors and use the GramSchmidt
orthonormalization process in each eigenspace. The vectors obtained in
this way will still be eigenvectors since they are linear combinations of
eigenvectors corresponding to the same eigenvalue. Vectors from different
eigenspaces will be orthogonal because of Theorem 9.3. Since eigenspaces
are seldom of very high dimensions, the amount of work involved in applying
the GramSchmidt process is usually quite nominal.
We now give several examples to illustrate the computational procedures
and the various diagonalization theorems. Remember that these examples
are contrived so that the characteristic equation can easily be solved. Ran
domly given examples of high order are very likely to result in vexingly
difficult characteristic equations.
12  The Computational Processes 215
Example 1. A real symmetric matrix with distinct eigenvalues. Let
"12 0"
A = 2 22
02 3_
We first determine the characteristic matrix,
C(x) =
"1  X
2
"
2
2 — x
2
2
3 *
and then the characteristic polynomial,
f{x) = det C(x) = x s + 6x*  3x  10 = (x + l)(x  2){x  5).
The eigenvalues are X 1 = — 1, A 2 = 2, A 3 = 5.
Solving the equations C{X t )X = we obtain the eigenvectors a x =
(2, 2, 1), a 2 = (2, 1, 2), a 3 = (1, 2, 2). Theorem 9.3 assures us that
these eigenvectors are orthogonal, and upon checking we see that they are.
Normalizing them, we obtain the orthonormal basis
X={£ 1 = (2, 2, 1), £ 2 = { (2, 1, 2), £ 3 = i(l, 2, 2)}.
The orthogonal matrix of transition is
"2 2 1"
P = i 2 1 2
1 2 2_
Example 2. A real symmetric matrix with repeated eigenvalues. Let
"5 2 2"
^ = 2 24
2 4 2.
The corresponding characteristic matrix is
"5  x 2 2
C(x) = 2 2 — x —4
2 4 2x
216 Orthogonal and Unitary Transformations, Normal Matrices  V
and the characteristic polynomial is
f(x) = x z + 9x 2  108 = 0 + 3)(a?  6) 2 .
The eigenvalues are A x = — 3, A 2 = A 3 = 6.
Corresponding to X r — —3, we obtain the eigenvector a x = (1, —2, —2).
For A 2 = A 3 = 6 we find that the eigenspace S(6) is of dimension 2 and
is the set of all solutions of the equation
Thus S(6) has the basis {(2, 1, 0), (2, 0, 1)}. We can now apply the Gram
Schmidt process to obtain the orthonormal basis
±(2, 1,0), ^=(2, 4, 5)).
Again, by Theorem 9.3 we are assured that a x is orthogonal to all vectors
in S(6), and to these vectors in particular. Thus,
X = f 1 (1, 2, 2), 4= (2, 1, 0), 7= (2, 4, 5)
U V5 3^5
is an orthonormal basis of eigenvectors. The orthogonal matrix of transition
is
~ 1
3
P =
2
2
V5
3V5
1
4
V5
3V5
5
V5_
It is worth noting that, whereas the eigenvector corresponding to an
eigenvalue of multiplicity 1 is unique up to a factor of absolute value 1,
the orthonormal basis of the eigenspace corresponding to a multiple eigen
value is not unique. In this example, any vector orthogonal to (1, —2, —2)
must be in S(6). Thus {£(2, 2, 1), £(2, 1,2)} would be another choice
for an orthonormal basis for S(6). It happens to result in a slightly simpler
orthogonal matrix of transition (in this case a matrix over the rational
numbers.)
Example 3. A Hermitian matrix. Let
2 I 1
A =
1 +/
12  The Computational Processes
Then
C(x) =
111
2 — x 1 — f
1 + i 3 — x
and/ (x) = x 2 — 5x + 4 = (a? — l)(a? — 4) = is the characteristic equation.
The eigenvalues are A x = 1 and A 2 = 4. (The example is contrived so that the
eigenvalues are rational, but the fact that they are real is assured by Theorem
10.1.) Corresponding to X x = 1 we obtain the normalized eigenvector
£ x = —j= (—1 + /, 1), and corresponding to A 2 = 4 we obtain the normalized
v3 j
eigenvector £ 2 = 7= (1,1+ 0 The unitary matrix of transition is
V3
V3
■1 + i
1
Example 4. An orthogonal matrix. Let
" 1 2
A=i 2 1
2 2
1
1 + iJ
This orthogonal matrix is real but not symmetric. Therefore, it is unitary
similar to a diagonal matrix but it is not orthogonal similar to a diagonal
matrix. We have
~ 2 2
C(x) =
A. Jl <*•
and, hence, —a? 3 + fa? 2 — fa? + 1 = — (x — l)(a? 2 + fa? + 1) = is the
characteristic equation. Notice that the real eigenvalues of an orthogonal
matrix are particularly easy to find since they must be of absolute value 1 .
I+2V2/ J „ I2V2/
The eigenvalues are / x = 1 , A 2 = , and x 3 = .
3 j 3
The corresponding normalized eigenvectors are  x = — ?=(1, —1,0), £ 2 =
V2
(1, 1, V2/), and  3 = (1, 1, — V2/) Thus, the unitary matrix of transition
is
£/ =
' 1/V2 i I
1/V2 I i
ify/2 1/V2.
218
Orthogonal and Unitary Transformations, Normal Matrices  V
EXERCISES
1. Apply the computational methods outlined in this section to obtain the
orthogonal or unitary matrices of transition to diagonalize each of the normal
matrices given in Exercises 1 of Sections 8, 10, and 11.
2. Carry out the program outlined in Exercises 12 through 15 of Section 11.
Consider the orthogonal linear transformation a represented by the orthogonal
matrix
A =
Find an orthonormal basis of eigenvectors of a + a*. Find the representation
of a with respect to this basis. Since a + o* has one eigenvalue of multiplicity 2,
the pairing described in Exercise 14 of Section 11 is not necessary. If a + a* had
an eigenvalue of multiplicity 4 or more, such a pairing would be required to obtain
the desired form.
chapter
VI
Selected
applications
of linear
algebra
In general, the application of any mathematical theory to any realistic
problem requires constructing a model of the problem in mathematical
terminology. How each concept in the model corresponds to a concept
in the problem requires understanding of both areas on the part of the person
making the application. If the problem is physical, he must understand the
physical facts that are to be related. He must also understand how the
mathematical concepts are related so that he can establish a correspondence
between the physical concepts and the mathematical concepts.
If this correspondence has been established in a meaningful way, pre
sumably the conclusions in the mathematical model will also have physical
meaning. If it were not for this aspect of the use of mathematical models,
mathematics could make little contribution to the problem for it could
otherwise not reveal any fact or conclusion not already known. The useful
ness of the model depends on how removed from obvious the conclusions are,
and how experience verifies the validity of the conclusions.
It must be emphasized that there is no hope of making any meaningful
numerical computations until the model has been constructed and under
stood. Anyone who attempts to apply a 'mathematical theory to a real
problem without understanding of the model faces the danger of making
inappropriate applications, or the restriction of doing only what someone
who does understand has instructed him to do. Too many students limit
their aims to remembering a sequence of steps that "give the answer" instead
of understanding the basic principles.
In the applications chosen for illustration here it is not possible to devote
more than token attention to the concepts in the field of the application.
For more details reference will have to be made to other sources. We do
identify the connection between the concept in the application and the con
cept in the model. Considerable attention is given to the construction of
219
220 Selected Applications of Linear Algebra  VI
complete and coherent models. In some cases the model already exists in
the material that has been developed in the first five chapters. This is true
to a large extent of the applications to geometry, communication theory,
differential equations, and small oscillations. In other cases, extensive
portions of the models must be constructed here. This has been necessary
for the applications to linear inequalities, linear programming, and repre
sentation theory.
1 I Vector Geometry
This section requires Chapter I, the first eight sections of Chapter II, and
the first four sections of Chapter IV for background.
We have already used the geometric interpretation of vectors to give the
concepts we were discussing a reality. We now develop this interpretation
in more detail. In doing this we find that the vector space concepts are
powerful tools in geometry and the geometric concepts are suggestive models
for corresponding facts about vector spaces.
We use vector algebra to construct an algebraic model for geometry.
In our imagination we identify a point P with a vector a from the origin to P.
a is called the position vector of P. In this way we establish a onetoone
correspondence between points and vectors. The correspondence will depend
on the choice of the origin. If a new origin is chosen, there will result a
different onetoone correspondence between points and vectors. The type of
geometry that is described by the model depends on the type of algebraic
structure that is given to the model. It is not our purpose here to get involved
in the details of various types of geometry. We shall be more concerned
with the ways geometric concepts can be identified with algebraic concepts.
Let V be a vector space of dimension n. We call a subspace S of dimension
1 a straight line through the origin. In the familiar model we have in mind
there are straight lines which do not pass through the origin, so this definition
must be generalized. A straight line is a set L of the form
L = a + S (1.1)
where a is a fixed vector and S is a subspace of dimension 1 . We describe
this situation by saying that S is a line through the origin and a displaces
S "parallel" to itself to a new position.
In general, a linear manifold or flat is a set L of the form
L = a + S (1.2)
where a is a fixed vector and S is a subspace. If S is of dimension r, we say
the linear manifold is of dimension r. A point is a linear manifold of dimen
sion 0, a line is of dimension 1 , a plane is of dimension 2, and a hyperplane
is a linear manifold of dimension n — 1 .
1 I Vector Geometry 221
Let V be the dual space of V and let S 1 be the annihilator of S. For every
<j> e S L we have
cf>(L) = 0(a) + <f>(S) = 0(a). (1.3)
On the other hand, let /? be any point in V for which <f>(fi) = 0(a) for all
e S 1 . Then 0(/9  a) = 0(#)  0(a) = so that  a e S; that is,
/? e a + S = L This means that S is identified by /3 as well as by a; that is,
L is determined by S and any vector in it.
Let L = a + S be of dimension r. Then S 1  is of dimension n — r. Let
{0 l5 . . . , 4> n _ r } be a basis of S 1 . Then
&0) = 4>M + <k( s ) = &(«) = c i (14)
for /= 1 ,...,« — r. Then /? £ L if and only if 0;(j#) = c* for i = 1 , . . . ,
n — r. Thus a linear manifold is determined by giving these n — r con
ditions, known as linear conditions. The linear manifold L is of dimension
r if and only if the n — r linear conditions are independent.
Two linear manifolds L x = a x + S ± and L 2 = a 2 + S 2 are said to be parallel
if and only if either S x c: S 2 or S 2 «= S x . If l^ and L 2 are of the same dimension,
they are parallel if and only if S x = S 2 .
Let Lj and L 2 be parallel and, to be definite, let us take S x c S 2 . Suppose
L x and L 2 have a point /8 = a x + #! = a 2 + c 2 in common. Then a x =
a 2 + (^2 — ^i) 6 a 2 + S 2 . Hence > ^ c f2 Thus, if two parallel linear
manifolds have a point in common, one is a subset of the other.
Let L = a + S be a linear manifold and let {a l5 . . . , a r } be a basis for S.
Then every vector /5 £ L can be written in the form
/? = a + f]^! + ' • • + *r a r (15)
As the t u ... , t r run through all values in the field F, p runs through the
linear manifold L For this reason (1.5) is called a parametric representation
Fie. 3
222 Selected Applications of Linear Algebra  VI
of the linear manifold L. Since a and the basis vectors are subject to a
wide variety of choices, there is no unique parametric representation.
Example. Let a = (1, 2, 3) and let S be a subspace of dimension 1 with
basis {(2, —1, 1)}. Then f = (x u x 2 , x 3 ) e a + S must satisfy the conditions
(x 1 ,x 2 ,x 3 ) = (1,2, 3) + t(2, 1,1).
In analytic geometry these conditions are usually written in the form
x x = 1 + 2/
x 3 = 3 + t,
the conventional or extended form of the parametric equations of a line.
The annihilator of S in this case has the basis {[1 2 0], [0 1 1]}. The
equations of the line a + S are then
x x + 2x 2 + 0% = 1 • 1 + 2 • 2 + = 5,
0*! + x 2 + x z = + 1 • 2 + 1 • 3 = 5.
With a little practice, vector methods are more intuitive and easier than the
methods of analytic geometry.
Suppose we wish to find out whether two lines are coplanar, or whether
they intersect. Let L, = a x + S x and L 2 = a 2 + S 2 be two given lines.
Then S x + S 2 is the smallest subspace parallel to both L x and L 2 . S x + S 2
is of dimension 2 unless S x = S 2 , a special case which is easy to handle.
Thus, assume S x + S 2 is of dimension 2. a x + S t + S 2 is a plane containing
Li and parallel to L 2 . Thus L x and L 2 are coplanar and L x intersects L 2 if
and only if <x 2 e a x + S 2 + S 2 . To determine this find the annihilator of
S x + S 2 . As in (1.3) oc a e <Xi + S x + S 2 if and only if (S x + Sa) 1 has the
same effect on both <x x and a 2 .
Example. Let L x = (1, 2, 3) + t(2, 1,1) and then let L 2 = (1, 1, 0)
+ 5(1,0,2). Then S x + S 2 = <(2, 1, 1), (1, 0, 2)) and (S 1 + S 2 y =
<[2 3  1]>. Since [2 3  1](1, 2, 3) = 5 and [2 3  1](1, 1, 0) = 5, the
lines ^ and L 2 both lie in the plane M = (1, 2, 3) + <(2, 1,1), (1,0, 2)>.
We can easily find the intersection of L x and L 2 . Since S x is a proper sub
space of S x + S 2 , (S x + Sjj) 1 is a proper subspace of S^. In this case the
difference in dimension is 1 , so we can find one linearly independent functional
in S/ that is not in (S x + S^ 1 , for example, [0 1 1]. Then a point of L 2
is in L x if and only if [0 1 1] has the same effect on it as it has on (1 , 2, 3).
Since [0 1 1](1,2, 3) = 5 and [0 1 1]{(1, 1, 0) + 5(1, 0, 2)} = 1 + 2s,
we see that 5 = 2. It is easily verified that (1 , 1 , 0) + 2(1 , 0, 2) = (3, 1 , 4) =
(1, 2, 3) + (2, 1, 1) is in both L x and L 2 .
1 I Vector Geometry 223
An important problem that must be considered is that of finding the
smallest linear manifold containing a given set of points. Let {P , P lf . . . , P n }
be a given set of points and let {<x , a 1? . . . , a B } be the corresponding set of
position vectors. For the sake of providing a geometric interpretation of the
algebra, we shall speak as though the linear manifold containing the set of
points is the same as the linear manifold containing the set of vectors.
A linear manifold containing these vectors must be of the form L = a + S
where S is a subspace. Since a can be any vector in L, we may as well take
L = a + S. Since a* — a e S, S contains {a! — a , . . . , a r — a }. On the
other hand, if S contains {a t  oc , . . . , a r  a }, then a + S will contain
{oc , a x , . . . , a r }. Thus a + <a x  a , . . . , a r  a > is the smallest linear
manifold containing {oc , . . . , a r }. If the {<x , . . . , a r } are given arbitrarily,
there is no assurance that K  <x , . . . , <x r  a } will be a linearly inde
pendent set. If it is, the linear manifold L will be of dimension r. In general,
two points determine a line, three points determine a plane, and n points
determine a hyperplane.
An arbitrary vector p will be in L if and only if it can be represented in the
form
= a + ^(a,  a ) + f 2 (a 2  a ) + • • • + t r (* r  a ). (1.6)
By setting 1 — h — ■ • •  t r = t , this expression can be written in the form
= t CL + fiCXi + • • • + t r cf. r , (1.7)
where
*0+fi+' + fr=l (1.8)
It is easily seen that (1.8) is a necessary and sufficient condition on the t t
in order that a linear combination of the form (1.7) lie in L
We should also like to be able to determine whether the linear manifold
generated by {a , a l9 . . . , a r } is of dimension r or of a lower dimension.
For example, when are three points colinear? The dimension of L is less
than r if and only if {a x  a , . . . , a r  a } is a linearly dependent set;
that is, if there exists a nontrivial linear relation of the form
Cx(a,  a ) + • • • + c r (a r  a ) = 0. (1.9)
This, in turn, can be written in the form
c a + c^ + • • • + c r a r = 0, (1.10)
where
Cb + Cx + '' + c^O. 010
It is an easy computational problem to determine the c it if a nonzero
solution exists. If a< is represented by (a u , a ti , . . . , a nj ), then (1.10) becomes
equivalent to the system of n equations
a i0 c + c a ci + • • ' + a ir c r = 0, /=!,...,«. (1.12)
224 Selected Applications of Linear Algebra  VI
These equations together with (1.11) form a system of n + 1 homogeneous
linear equations to solve". As a system, they are equivalent to determining
whether the set {(1, a Xj , a 2i , . . . , a nj ) \j = 0, 1, . . . , r} is linearly dependent.
Geometrically, the pair of conditions (1.7), (1.8) and the pair of conditions
(1.10), (1.1 1) are independent of the choice of an origin. Suppose the points
P , P lt . . . , P r are identified with position vectors from some other point.
For example, let be the origin and let 0' be a new choice for an origin.
If a' is the position vector of 0' with reference to the old origin, then
a i = a i — a ' (113)
is the position vector of P t relative to 0'. If in (1.7) is the position vector
of B relative to and 0' = — a' is the position vector of B relative to 0',
then (1.7) takes the form
r
ft' = /3 — a' = 2 t k <x. k — a'
fc=0
r
k=0
=i*x (i.7)'
fc=0
Also, (1.10) takes the form
r r
2 c kK = 2 c fc( a fc  a')
fc=0 k=0
r r
= 2 c k*k  2 c fc a '
fc=0 fc=0
=i% = o. (i.io)'
k=0
Since the pair of conditions (1.7), (1.8) is related to a geometric property,
it should be expected that if a pair of conditions like (1.7), (1.8) hold with one
choice of an origin, then a pair of similar conditions should hold for another
choice. The real importance of the observation above is that the new pair
of conditions involve the same coefficients.
If (1.7) holds subject to (1.8), we call /S an affine combination of {<x ,
a l5 . . . , a r }. If (1.10) holds with coefficients not all zero subject to (1.11), we
say that {a , a l5 . . . , a r } is an affinely dependent set. The concepts of affine
combinations, affine dependence, and affine independence are related to each
other in much the same way that linear combinations, linear dependence,
and linear independence are related to each other. For example, the affine
combination (1.7) is unique if and only if the set {a , a 1} . . . , a r } is affinely
independent. The set {a , a 1? . . . , a r } is affinely dependent if and only if one
vector is an affine combination of the preceding vectors.
1 I Vector Geometry 225
Affine geometry is important, but its study is much neglected in American
high schools and universities. The reason for this is primarily that it is diffi
cult to study affine geometry without using linear algebra as a tool. A good
picture of what concepts are involved in affine geometry can be obtained by
considering Euclidean geometry in the plane or 3space in which we "forget"
about distance. In Euclidean geometry we study properties that are unchanged
by rigid motions. In affine geometry we allow nonrigid motions, provided
they preserve intersection, parallelism, and colinearity. If an origin is intro
duced, a rigid motion can be represented as an orthogonal transformation
followed by a translation. An affine transformation can be represented by a
nonsingular linear transformation followed by a translation. If we accept
this assertion, it is easy to see that affine combinations are preserved under
affine transformations.
Theorem 1.1. Let {L A :X g A} be any collection of linear manifolds in V.
Either n AeA L A is empty or it is a linear manifold.
proof. If r\te)\x is not empty, let a e t^^ eA L x . Since a £ L x and L A is a
linear manifold, i x = oc + S x where S A is a subspace of V. Then n A6A L A =
n A6A( a o + S x ) = a + n AeA S A . Since n AeA S A is a subspace, n xeA L x is a linear
manifold. □
Definition. Given any nonempty subset S of V, the affine closure of S is the
smallest linear manifold containing S. In view of Theorem 1.1, the affine
closure of S is the intersection of all linear manifolds containing S, and this
shows that a smallest linear manifold containing S actually exists. The
affine closure of S is denoted by A(S).
Theorem 1.2. Let S be any subset of V. Let S be the set of all affine com
binations of finite subsets ofS. Then S is a linear manifold.
proof. Let {/3 , j8j, . . . , p k } be a finite subset of S. Each & is of the form
Tj
Pi = 2* X i) IX ii
where
r
2 xa = i
1=0
and each a fj e S. Then for any /S of the form
/* = i 'A
3=0
and
i u = i,
i=0
226 Selected Applications of Linear Algebra  VI
we have
P = 2 M 2 x u*ii)
3=0 \ i=0 /
and
= 22 x a x Pi
3=0 i=0
2 2 x n f i = 2 h 2
3=0 i=0 j'=0 i=0
= 2',
i=o
= 1.
Thus /? gS. This shows that S is closed under affine combinations, that is,
an affine combination of a finite number of elements in S is also in S.
The observation of the previous paragraph will allow us to conclude that S
is a linear manifold. Let <x be any fixed element in S. Let S — <x denote the
set of elements of the form a — a where a e S. If {j8i, 8 8 , . . . , ft.} is a finite
subset of S — <x , where ft = a* — <x , then 2* =1 c ift = 2*=i c * a i ~ 2*=i
*i«o = 2io_ c * a *  a o where c = 1  2ti c i Thus 2*=o c i*i eS * nd
2i=i C A eS — a . This shows that S — a is a subspace. Hence, S =
a + (S — a ) is a linear manifold. □
Theorem 1.3. The affine closure ofS is the set of all affine combinations of
finite subsets ofS.
Since S is a linear manifold containing S, A(S) <= S. On the other hand,
A(S) is a linear manifold containing S and, hence, A(S) contains all affine
combinations of elements in S. Thus S ez A(S). This shows that S = A(S). □
Theorem 1.4. Let l x = a x + S x awd /e/ L 2 = a 2 + S 2 . 77j£?n A(Lx U L 2 ) =
«i + < a 2 — a i) + $i + S 2 .
proof. Clearly, L x U L 2 c Kl + <a 2  a x ) + S x + S 2 and a x + <a 2 
a i) + Si + S 2 is a linear manifold containing L x U L 2 . Since <x x e Lj U L 2 <=
A(L X u La), A^ u L 2 ) is of the form a x + S where S is a subspace. And since
a 2 6L 1 uL 2 ca 1 + S,a 1 + S = a 2 + S. Thus a 2  a x 6 S. Since a x + S x =
L x c= L x U L 2 c ai + S, S x cr S. Similarly, S 2 c S. Thus (<x 2  clj) + S x +
S 2 c S, and a x + <a 2  a x ) + S x + S 2 c ai + S = A^ U L 2 ). This shows
that A(L X V L 2 ) = a x + <a 2 — a x > + S x + S 2 . D
Theorem 1.5. Let L x = a x + S x a/w/ fef L 2 = a 2 + S 2 6e /wear manifolds.
L x n L 2 = if and only if a 2 — a x ^ S x + S 2 .
proof. If L x n L 2 is not empty, let a e L x n L 2 . Then L x = a + S t =
a x + S x and L 2 = a + S 2 = a 2 + S 2 . Thus a„  a x G S x and a 2 — a e S 2 .
I I Vector Geometry 227
Hence <x 2 — <x x = (ot — a x ) + (a 2 — <*o) e S x + S 2 . Conversely, if a 2 — a x e
S x + S 2 then a 2 — a x = y t + y 2 where y x e S x and y 2 e S 2 . Thus a x + y y =
a 2 — y 2 e ( a i + $i) n (*2 + S a ). D
Corollary 1.6. L x C\ L 2 = j/ a«J o«/y */ dim A(L 1 U L 2 ) = dim (S x +
S 2 ) + 1. 7/L x HL^D, dim A(L X U L 2 ) = dim (S x + S 2 ). a
We now wish to introduce the idea of betweenness. This is a concept
closely tied to the real numbers, so we assume for the rest of this section that
F is the field of real numbers or a subfield of the real numbers.
Let ax and <x 2 be any two vectors. We have seen that every vector ft in the
line generated by a x and <x 2 can be written in the form ft = ^aj + / 2 a 2 ,
where t x + t 2 = 1 . This is equivalent to the form
p = (1  0«i + '«2 (114)
For / = 0, ft = « ls and for t = 1, /5 = oc 2 . We say that /S is between
a t and a 2 if and only if t is between and 1. The line segment joining a x
and <x 2 consists of all points of the form ^oc! + / 2 a 2 where t x + t 2 = 1 and
t x > and r 2 > 0. A subset C of V is said to be convex if, whenever two
points are in C, every point of the line segment joining them is also in C.
Clearly, the space V itself is convex, and every subspace of V is convex.
Exercise 1 1 of Chapter IV4 amounts to showing that the two sides of a hyper
plane are convex.
Theorem 1.7. The intersection of any number of convex sets is convex.
proof. Let {C a } AgA be a collection of convex sets. If a x and a 2 are in
n AeA C A , then for each X both a x and a 2 are in C x . Since C x is convex, the
segment joining a x and <x 2 is in C x . Thus the segment is in n XeA C x and the
intersection is convex. □
As a slight generalization of the expression we gave for the line segment
joining two points, we define a convex linear combination of elements of a
subset S to be a vector ft expressible in the form
ft = ^o^ + ? 2 a 2 + • • • + t r a. r , (1.15)
where
*i + h + ••• + t r = 1, u >0, (1.16)
and {a l9 a 2 , . . . , a r } is a finite subset of S. If S is a finite subset, a useful and
informative picture of the situation is formed in the following way: Imagine
the points of S to be contained in a plane or 3dimensional space. The set of
convex linear combinations is then a polygon (in the plane) or polyhedron
in which the corners are points of S. Those points of S which are not at
the corners are contained in the edges, faces, or interior of the polygon or
228 Selected Applications of Linear Algebra  VI
polyhedron. Tt is the purpose of Theorem 1.9 to prove that this is a depend
able picture.
Theorem 1.8. A set C is convex if and only if every convex linear combina
tion of vectors in C is in C.
proof. If every convex linear combination of vectors in C is in C, then,
in particular, every line segment joining a pair of vectors in C is in C. Thus
C is convex.
On the other hand, assume C is convex and let (5 = 2*=i t^ be a convex
linear combination of vectors in C. For r = 1, /? = oc x 6 C; and for r = 2,
is on the line segment joining o^ and a 2 hence (3 e C. Assume that a convex
linear combination involving fewer than r elements of C is in C. We can
assume that t r ^ 1, for otherwise /3 = a r g C. Then for each /, (1 — ? r )a^ +
La, 6 C and
r1
/? = 2 T^~ { (1  ^i + *r«r} (117)
i=l 1 — t r
is a convex linear combination of r — 1 elements of C, and is therefore
in C. □
Let S be any subset of V. The convex hull H(S) of S is the smallest convex
set containing S. Since V is a convex set containing S, and the intersection
of all convex sets containing S is a convex set containing S, such a smallest
set always exists.
Theorem 1.9. The convex hull of a set S is the set of all convex linear
combinations of vectors in S.
proof. Let T be the set of all convex linear combinations of vectors
in S. Clearly, S c Tand, since H(S) contains all convex linear combinations
of vectors in S, T <= H(S). Thus the theorem will be established if we show
that T is convex. Let a, ft eT. These vectors can be expressed in the form
a = 2 hvi.
u > o,
a^eS,
r
= 2 s i*i>
r
2 s * = !>
s t > 0,
a^eS,
i=l i=l
where both expressions involve the same finite set of elements of S. This
can be done by adjoining, where necessary, some terms with zero coefficients.
Then for < / < 1 ,
(1 _ Ooc + ,p = j {(1  t )t t + ttja,.
«=i
Since (1  t)t t + ^ > and 2l=i (0 ~ 0'» + ts t } = (1  + t = 1,
(1 — r)a + r/5 6 T and T is a convex set. Thus T = H(S). □
2  Finite Cones and Linear Inequalities 229
BIBLIOGRAPHICAL NOTES
The connections between linear algebra and geometry appear to some extent in almost
all expository material on matrix theory. For an excellent classical treatment see M.
Bocher, Introduction to Higher Algebra. For an elegant modern treatment see R. W.
Gruenberg and A. J. Weir, Linear Geometry. A very readable exposition that starts from
first principles and treats some classical problems with modern tools is available in N. H.
Kuiper, Linear Algebra and Geometry.
EXERCISES
1. For each of the following linear manifolds, write down a parametric repre
sentation and a determining set of linear conditions :
(3) L x = (1,0,1) + <(1,1,1), (2, 1,0)>.
(2) L 2 = (1, 2, 2) + <(2, 1, 2), (2, 2, 1)>.
(3) L 3 = (1, 1, 1, 2) + <(0, 1, 0, 1), (2, 1, 2, 3)>.
2. For L x and L 2 given in Exercise 1 , find a parametric representation and linear
conditions for L x n L 2 .
3. In R 3 find the smallest linear manifold L containing {(2, 1, 2), (2, 2, 1),
( — 1,1,2")}. Show that L is parallel to L x n L 2 , where L x and L 2 are given in Exercise 1.
4. Determine whether (0,0) is in the convex hull of S = {(1, 1), (6, 7), (5, 6)}.
(This can be determined reasonably well by careful plotting on a coordinate system.
At least it can be done with sufficient accuracy to make a guess which can be verified.
For higher dimensions the use of plotting points is too difficult and inaccurate. An
effective method is given in Section 3.)
5. Determine whether (0, 0, 0) is in the convex hull of T = {(6, —5, —2),
(3, 8, 6), (4, 8, 5), (9, 2, 8), (7, 2, 8), (5, 5, 1)}.
6. Show that the intersection of two linear manifolds is either empty or a linear
manifold.
7. If L ± and L 2 are linear manifolds, the join of L x and L 2 is the smallest linear
manifold containing both L x and L 2 which we denote by L x J L 2 . If L r = a x + S x and
L 2 = <x 2 + S 2 , show that the join of L x and L 2 is a x + <a 2 — a x > + S 1 + S 2 .
8. Let ii = ai + S x and L 2 = a 2 + S 2 . Show that if L x n L 2 is not empty, then
L x J L 2 = 04 + S x + S 2 .
9. Let ^ = 04 + S x and L 2 = <x 2 + S 2 . Show that if L x n L 2 is empty, then
Lj J L 2 5* <x x + S x + S 2 , that is, <x 2 — a x ^ S x + S 2 .
10. Show that dim L x J L 2 = dim (S x + S 2 ) if L x n L 2 ^ and dim L x J L 2 =
dim (S x + S 2 ) + 1 if l^ n L 2 = 0.
2 I Finite Cones and Linear Inequalities
This section requires Section 1 for background and, of course, those sec
tions required for Section 1. Although some material is independently
developed here, Section 10 of Chapter II would also be helpful.
230 Selected Applications of Linear Algebra  VI
In this section and the following section we assume that F is the field of
real numbers R, or a subfield of R.
If a set is closed under multiplication by nonnegative scalars, it is called
a cone. This is in analogy with the familiar cones of elementary geometry
with vertex at the origin which contain with any point not at the vertex
all points on the same halfline from the vertex through the point. If the
cone is also closed under addition, it is called a convex cone. It is easily seen
that a convex cone is a convex set.
If C is a convex cone and there exists a finite set of vectors {a x , . . . , a. P }
in C such that every vector in C can be represented as a linear combination
of the <Xj with nonnegative coefficients, a nonnegative linear combination,
we call {oc x , . . . , a^,} the generators of C and call C a. finite cone. The cone
generated by a single nonzero vector is called a halfline. A dependable
picture of a finite cone is formed by considering the halflines formed by each
of the generators as constituting an edge of a pointed cone as in Fig. 4. By
considering a solid circular cone in R 3 it should be clear that there are convex
cones that are not finite. A finite cone is the convex hull of a finite number
of halflines.
Let S be the largest subspace contained in C. If S = {0}, then S contains
no line through the origin. In this case we say that C is pointed. If S is of
dimension 1 , then C is wedge shaped with S forming the edge of the wedge.
Given any subset W <= V, let W+ denote the set of all linear functionals
that take on nonnegative values for all a e W; that is, W+ = {<f>  (fxx >
for all a £ W}. VV+ is closed under nonnegative linear combinations and
is a convex cone in V. W+ is called the dual cone or polar cone of W. Similarly,
if VV c: V, then W+ is the set of all vectors which have nonnegative values for
Fig. 4
2  Finite Cones and Linear Inequalities 231
all linear functionals in W. In this case, too, W+ is called the dual cone of W.
For the dual of the dual (W+)+ we write VV++.
Theorem 2.1. (1) If W x c VV 2 , then W+ => VV+.
(2) (W 1 + W 2 )+ = W+ n W+ */0 e W x n W 2 .
(3) W+ + W+ c (W x n W 2 )+.
proof. (1) is obvious.
(2) If <f>eW+nW+, then for all a = a x + a 2 where a x e W x and
a 2 e W 2 we have cfxx = fa + fa > 0. Hence, VV+ n VV+ c (W x + W 2 )+.
On the other hand, ^cW^ W 2 so that W+ ^ (W 1 + W 2 )+. Similarly,
W+ => (W x + W 2 )+. Hence, W+ n W+ => (Wj + W 2 )+. It follows then
that VV+ n VV+ = (VV X + W 2 )+.
(3) Wj 3 W t n VV 2 so that VV+ c (^ n W 2 )+. Similarly, W+ c
(W x n W 2 )+. It then follows that W+ + W+ c (W 1 n W 2 )+. n
Theorem 2.2. W <= W++ an^ VV+ = VV+++.
proof. Let W <= V. If a e W, then 0a > for all <£ e W+. This means
that W c W++. It then follows that W+ c (W+)++ = W+++ On the other
hand from Proposition 2.1 we have VV+ => (W++)+ = VV+++. Thus W+ =
W+++. The situation is the same for W c V. n
A cone C is said to be reflexive if C = C++.
Theorem 2.3. A cone is reflexive if and only if it is the dual cone of a set
in the dual space.
proof. Suppose C is reflexive. Then C = C++ is the dual cone of C+.
On the other hand, if C is the dual cone ofWcf, then C = W+ = W+++ =
C++ and C is reflexive. □
The dual cone of a finite cone is called a polyhedral cone. If C is a finite
.A.
cone in V generated by the finite set G = {fa, . . . , <f) Q }, then C + = D =
{a  faa. > for all fa e G}. A dependable picture of a polyhedral cone
can be formed by considering a finite cone, for we soon show that the two
types of cones are equivalent. Each face of the cone is a part of one of the
hyperplanes {a j ^a = 0}, and the cone is on the positive side of each of
these hyperplanes. In a finite cone the emphasis is on the edges as generating
the cone; in a polyhedral cone the emphasis is on the faces as bounding
the cone.
Theorem 2.4. Let o be a linear transformation of U into V.IfC is a finite
cone in U, then c(C) is a finite cone. IfD is a polyhedral cone in V, then a^D)
is a polyhedral cone.
proof. If {a x , . . . , a p } generates C, then {ofa), . . . , <r(a p )} generates
c(C). Let D be a polyhedral cone dual to the finite cone £ in V. The following
232 Selected Applications of Linear Algebra  VI
statements are equivalent: <x e cr 1 (D); cr(a) e D; \pa{<x) > for all ip e £;
<Ky) a > for all tp g £; ae (cr(E))+. Thus (y~ x {D) is dual to the finite cone
(x(E) in and is therefore polyhedral. □
Theorem 2.5. The sum of a finite number of finite cones is a finite cone
and the intersection of a finite number of polyhedral cones is a polyhedral cone.
proof. The first assertion of the theorem is obvious. Let D 1} . . . , D r
be polyhedral cones, and let C l5 . . . , C r be the finite cones of which they
are the duals. Then C x + • • • + C r is a finite cone, and by Theorem 2.1
D x n • • • C\D r = C+ n • • • nC+ = (C x + • • • + C r )+ is polyhedral. □
Theorem 2.6. Every finite cone is polyhedral.
proof. The theorem is obviously true in a vector space of dimension 1 .
Let dim V = n and assume the theorem is true in vector spaces of dimension
less than n.
Let A = {oc 1? . . . , a p } be a finite set generating the finite cone C. We
can assume that each a k t± 0. For each a. k let W k be a complementary sub
space of (a fc >; that is, V = W ft © (a. k ). Let iT k be the projection of V onto
W k along <a fc >. 7r fc (C) is a finite cone in W^. By the induction assumption it is
polyhedral since dim W k = n — 1. Then ir 1 (n k (C)) = C k is polyhedral by
Theorem 2.4. Since C <= C k for each A;, C is contained in the polyhedral cone
C,n nC p .
We must now show that if <x ^ C and C is not a halfline, then there is a
Cj such that <x £ C,. If not, then suppose a e C j for j = 1 , . . . , p. Then
77\,.(a ) e •nj(C) so that there is an a i e F such that a + a,a 3  = ]£?=i ^a^t where
6»j !> 0. We cannot obtain such an expression with a t < for then oc would
be in C. But we can modify these expressions step by step and remove all
the terms on the right sides.
Suppose b i:j = for / < k and j = 1, . . . ,p; that is, <x + a^ =
^,i=k^a a i This is already true for k = 1. Then
p
i=k+l
As before, we cannot have a k — b kk < 0. Set a k — b kk = a' k > 0. Then for
j j£ k we have
(l + M«o + ap, = I (b is + % fc ft W
\ a fc / i=fc+i\ a k J
Upon division by 1 + y we get expressions of the form
V
a + aja, = 2 &« a *> J = 1» 2, . . . , p,
2  Finite Cones and Linear Inequalities 233
with a] > and b'^ > 0. Continuing in this way we eventually get a +
c^ = with c, > for ally. This would imply a, = a fory = 1, . . . , p.
Thus C is generated by {— a }; that is, C is a halfline, which is polyhedral.
This would imply that C is polyhedral. If C is polyhedral there is nothing
to prove. If C is not halfline, the assumption that <x e C, for all j is un
tenable. But this would mean that C = C 2 n • ■ • n C p , in which case C is
polyhedral. □
Theorem 2.7. A polyhedral cone is finite.
proof. Let C = D+ be a polyhedral cone dual to the finite cone D.
We have just proven that a finite cone is polyhedral, so there is a finite cone
£ such that D = E+. But then £ is also polyhedral so that £ = E++ =
D+ = C. Since £ is finite, C is also. □
Although polyhedral cones and finite cones are identical, we retain both
terms and use whichever is most suitable for the point of view we wish to
emphasize. A large number of interesting and important results now follow
very easily. The sum and intersection of finite cones are finite cones. The
dual cone of a finite cone is finite. A finite cone is reflexive. Also, for finite
cones part (3) of Theorem 2.1 can be improved. If C x and C 2 are finite cones,
then (C, n C 2 )+ = (C++ n C++)+ = (C+ + C+)++ = C+ + C+.
Our purpose in introducing this discussion of finite cones was to obtain
some theorems about linear inequalities, so we now turn our attention to
that subject. The following theorem is nothing but a paraphrase of the
statement that a finite cone is reflexive.
Theorem 2.8. Let
a 11 x 1 + • • • + a ln x n >
(2.1)
fl«A + • • * + a m n x n >
be a system of linear inequalities. If
a x x x + • • • + a n x n >
is a linear inequality which is satisfied whenever the system (2.1) is satisfied,
then there exist nonnegative scalar s {y x , . . . , y m ) such that ^™ =1 y^ = a$
forj= I, ... ,n.
proof. Let (f> t be the linear functional represented by [a n • • • a in ],
and let <f> be the linear functional represented by [a x • • • a n ]. If I represented
by (x lt . . . ,x n ) satisfies the system (2.1), then I is in the cone C+ dual to the
234 Selected Applications of Linear Algebra  VI
finite cone C generated by {<f> x , . . . , <f> m }. Since (f>£ > for all £ e C+,
<f> g C++ = C. Thus there exist nonnegative y { such that <f> = ^,™ =1 Vi<f>i
The conclusion of the theorem then follows. D
Theorem 2.9. Let A = {<x l5 . . . , <x n } be a basis of the vector space U and
let P be the finite cone generated by A. Let a be a linear transformation of U
into V and let (S be a given vector in V. Then one and only one of the following
two alternatives holds: either
(1) there is a £ e P such that a{£) = jS, or
(2) there is a tp eV such that o(ip) e P+ and ipfi < 0.
proof. Suppose (1) and (2) are satisfied at the same time. Then
> ip(S = ipo(g) = 6(yi)£ > 0, which is a contradiction.
On the other hand, suppose (1) is not satisfied. Since P is a finite cone,
a(P) is a finite cone. The insolvability of (1) means that /? ^ o{P). Since
a(P) is also a polyhedral cone, there is a tp e V such that y>/3 < and tpo , {P) >
0. But then a(y)(P) > so that 6(rp) 6 P+. □
It is apparent that the assumption that A is a basis of U is not used in the
proof of Theorem 2.9. We wish, however, to translate this theorem into
matrix notation. If  is represented by X = (a; l5 . . . , x n ), then £ e P if
and only if each a^ > 0. To simplify notation we write "X > 0" to mean
each x t > 0, and we refer to P as the positive orthant. Since the generators
of P form a basis of U, the generators of P+ are the elements of the dual basis
A = {<Ai> • • • 5 <l>n} It thus turns out that P+ is the positive orthant of 0.
Let /? = {&, . . . , /8 m } be a basis of V and B = {&, . . . , fi m } the dual
basis in V. Let ,4 = [a tj ] represent a with respect to A and 6, 5 = (b lf . . . ,
b m ) represent /?, and Y = [y x • • • y m ] represent tp. Then a(ip) is represented
by YA and a(yi) e P+ if and only if YA > 0. In this notation Theorem 2.9
becomes
Theorem 2.10. One and only one of the following two alternatives holds:
either
(1) there is an X > such that AX = B, or
(2) there is a Y such that YA > and YB < 0. □
Rather than continue to make these translations we adopt notational
conventions which will make such translations more evident. We write
1 > to mean £ e P, £ > £ to mean £ — £ e P, <r(^) > to mean a(ip) e P+,
etc.
Theorem 2.11. With the notation of Theorem 2.9, let <f> be a linear func
tional in U, let g be an arbitrary scalar, and assume {3 £ a(P). Then one and
only one of the following two alternatives holds; either
(1) there is a  > such that <t(£) = /? and <f>£ > g, or,
(2) there is ip e V such that 6(ip) > <f> and xpfi < g.
2  Finite Cones and Linear Inequalities 235
proof. Suppose (1) and (2). are satisfied at the same time. Then
g > y)fi = ^cr(l) = o(y>)£ > <f>£ > g, which is a contradiction.
On the other hand, suppose (2) is not satisfied. We wish to find a f e P
satisfying the conditions er(£) = j3 and <£ > g at the same time. We have
seen before that vectors and linear transformations can be used to express
systems of equations as a single vector equation. A similar technique works
here.
Let U t = U © F be the set of all pairs (, x) where £ e U and xeF. U x
is made into a vector space over F by defining vector addition and scalar
multiplication according to the rules
(fi» *i) + (fa. *s) = (£i + ?2» *i + *«)> «(£» *) = ( fl f » «*)■
Let P be the set of all (£, a;) where I e P and a; ^> 0. It is easily seen that P
is a finite cone in U v In a similar way we construct the vector space V x =
V©F.
We then define S to be the mapping of U x into V x which maps (I, *)
onto 2(, a;) = (<r(!), </*£ — x ) Jt can De checked that 2 is linear. It is
now seen that (g,x)eP and E(£, a) = (8, g) are equivalent to the conditions
I e P, <r(£) = p, and 0£ = g + x £ g. ^
To use Theorem 2.9 we must describe U x and Vi and determine the adjoint
transformation E. It is not difficult to see that U © F is isomorphic to 1/ F
where ((£, y) is the linear functional defined by the formula (<f>, y)(f , a;) =
<£ + ya\ In a similar way V © F is isomorphic to V © F. Then 2(y>, y)
applied to (f , x) must have the effect
S(V, */(£ , *) = (v» y) 2 (£ » *)
= (y, *)(*(*), *f  *)
= y><T(g) + y(^f  *)
= #(v0£ + #1  y*
= (#(v) + y<f>)£  v x 
This means that S(y, y) = (ff(y>) + y<£, —y).
Now suppose there exist xp £ V and yeF for which S(y», y) g P + and
(V> #)(/*» g) = V>P + Vg < 0 Tnis is m tne form of condition (2) of Theorem
2.9 and we wish to show that it cannot hold. S(y, y) e P+ means <7(y) + y<f>^.
and — y > 0. If — y > 0, then ol— I > <f>. Since (2) of this theorem is
assumed not to hold this means I — J/5 > g, or tpp + yg > 0. If y = 0,
then a(\p) > 0. Since /8 e a(P) by assumption, yfi = yp + yg < would
contradict Theorem 2.9. Thus (2) of Theorem 2.9 cannot be satisfied. This
236 Selected Applications of Linear Algebra  VI
implies there is a (£, x)eP such that 2(£, x) — (<r(£). cf>£  x) = (ft, g),
which proves the theorem. □
Theorem 2.12. Let P x be the positive orthant in U generated by the basis
A = {a 1? . . . , a„} and P 2 the positive orthant in V generated by the basis
8 = {#i, . . . , /5 m }. Let a, /?, and <f> be given and assume /S e a(P 1 ) + P 2 . For
each scalar g one and only one of the following two alternatives holds, either
(1) there is a £ > such that cr(£) < /5 am/ </>£ > g, or
(2) ?/*ere is a ip > swc/i /Aa? ^(y) > </> a«J ipfi < g.
proof. Construct the vector space U © V and define the mapping S
of U © V into V by the rule
2(£, 77) = <r(£) + 77.
Then the condition E(£, rj) = ft with (£, 77) ^ is equivalent to ft — cr(£) =
t? > with £ > 0. Since Z(y>) = (d(y>)» v), the condition S(v>) > (</>, 0)
is equivalent to a(ip) > ip and ^ > 0. With this interpretation, the two
conditions of this theorem are equivalent to the corresponding conditions of
Theorem 2.11. □
Theorem 2.13. Let a, /S, and <f> be given and assume fi e a(U). For each
scalar g one and only one of the following two alternatives holds, either
(1) there is a £ e U such that <?(£) = (3 and </>£ > g, or
(2) there is a ip eV such that a(ip) = <f> and ipfi < g.
proof. Construct the vector space U © U and define the mapping 2
of U © U into V by the rule
S(&, *■) = *(£i)  *(&) = *(£i  l 2 ).
Let £ = £ x  £ 2 . Then the condition S^, £ 2 ) = with (£ 1? £ 2 ) > is
equivalent to <r(£) = /? with no other restriction on £ since every £ e U
can be represented in the form £ = £ x — £ 2 with £1 > and £ 2 > 0. Since
A. A.
S(v) = (ffCyO* — o'Cv))' the condition 2(^) > (</», —0) is equivalent to
a(ip) = cf>. With this interpretation, the two conditions of this theorem
are equivalent to the corresponding conditions of Theorem 2.11. □
Notice, in Theorems 2.1 1 , 2.12, and 2.13, how an inequality for one variable
corresponds to the condition that the other variable be nonnegative, while
an equation for one of the variables leaves the other variable unrestricted.
For example, we have the conditions a(£) = /S and ip e V in Theorem 2.11
replaced by o(g) < /S and ip > in Theorem 2.12.
Theorem 2.14. One and only one of the following two alternatives holds:
either
(1) there is a £ > such that a(£) = and <f>£ > 0, or
A.
(2) there is a ip eV such that 6{ip) > <j>.
2  Finite Cones and Linear Inequalities 237
proof. This theorem follows from Theorem 2.11 by taking p = and
g > 0. The assumption that £ <r(P) is then satisfied automatically and
the condition y>@ < g is not a restriction. □
Theorem 2.15. One and only one of the following two alternatives holds:
either
(1) there is a I > such that o(£) < (3, or
(2) there is a ip > rac/i that a (if) > an<5? y>/? < 0.
proof. This theorem follows from Theorem 2.12 by taking <j> = and
g < 0. In this case the assumption that e cr^) + P 2 is not satisfied
automatically. However, in Theorem 2.12 conditions (1) and (2) are sym
metric and this assumption could have been replaced by the dual assumption
that —cf>eP+ + a(— P+), and this assumption is satisfied. □
Theorem 2.16. One and only one of the following two alternatives holds:
either
(1) there is a I £ U such that o(£) = /?, or
(2) there is a xp £ V such that a(ip) = and tpfi < 0.
proof. This theorem follows from Theorem 2.13 by taking (f> = and
g < 0. Again, although the condition /? £ a(U) is not satisfied automatically,
the equally sufficient dual condition <j> £ cr(V) is satisfied. □
It is sometimes convenient to express condition (2) of Theorem 2.16 in a
slightly different form. It is equivalent to assert that there is a ip £ V such
that a(rp) = and ytfi = 1. If y satisfies condition (2), then — satisfies this
condition. In this form Theorem 2.16 is equivalent to Theorem 7.2 of
Chapter II, and identical to Theorem 5.5 of Chapter IV.
An application of these theorems to the problem of linear programming is
made in Section 3.
BIBLIOGRAPHICAL NOTES
An excellent expository treatment of this subject with numerous examples is given by
D. Gale, The Theory of Linear Economic Models. A number of expository and research
papers and an extensive bibliography are available in Kuhn and Tucker, Linear Inequalities
and Related Systems, Annals of Mathematics Studies, Study 38.
EXERCISES
1. In R 3 let C x be the finite cone generated by {(1, 1,0), (1,0, 1), (0, 1, 1)}.
Wrtie down the inequalities that characterize the polyhedral cone C+.
2. Find the linear functional which generate C+ as a finite cone.
238
Selected Applications of Linear Algebra  VI
3. In R 3 let C 2 be the finite cone generated by {(0, 1,1), (1, 1,0), (1,1, 1)}.
Write down a set of generators of C l + C 2 , where C x is the finite cone given in
Exercise 1.
4. Find a minimum set of generators of C x + C 2 , where C x and C 2 are the cones
given in Exercises 1 and 3.
5. Find the generators of C+ + C+, where C x and C 2 are the cones given in
Exercises 1 and 3.
6. Determine the generators of C x n C 2 .
"0 1 1"
7. Let A =
1 1
1
and
B =
Determine whether there is an X > such that AX = B. (Since the columns of
A are the generators of C 2 , this is a question of whether (1, 1, 0) is in C 2 . Why?)
8. Use Theorem 2.16 (or the matrix equivalent) to show that the following
system of equations has no solution :
2x x + 2x 2 = —1.
9. Use Theorem 2.9 (or 2.10) to show that the following system of equations
has no nonnegative solution :
10. Prove the following theorem: One and only one of the following two
alternatives holds : either
(1) there is a £ e U such that ct(£) > p, or
(2) there is a y> > such that ff(y) = and y>(? > 0.
1 1 . Use the previous exercise to show that the following system of inequalities
has no solution.
2x x + 2x 2 > 1.
12. A vector I is said to be positive if I = ^,f =1 x i^i where each x i > and
{£ x , . . . , £ n } is a basis generating the positive orthant. We use the notation f >
to denote the fact that I is positive. A vector I is said to be semipositive if I >
and ^#0. Use Theorem 2.11 to prove the following theorem. One and only one
of the following two alternatives holds : either
(1) there is a semipositive £ such that cr() = 0, or
(2) there is a y> e V such that 6(xp) is positive.
3  Linear Programming 239
13. Let W be a subspace of U, and let W 1 be the annihilator of W. Let l^, ... ,
%} be a basis of W 1 . Let V be any vector space of dimension r over the same
coefficient field, and let {j8 l5 . . . , /3 r } be a basis of V. Define the linear transformation
a of U into V by the rule, <r(f) = Jji *?;(£)&• Show that W is the kernel of <r.
Show that W^ = 6(V).
14. Show that if W is a subspace of U, then one and only one of the following
two alternatives holds: either
(1) there is a semipositive vector in W, or
(2) there is a positive linear functional in W 1  .
15. Use Theorem 2.12 to prove the following theorem. One and only one of the
following two alternatives holds : either
(1) there is a semipositive £ such that <r(£) < 0, or
(2) there is a y > such that 6(y>) > 0.
3 I Linear Programming
This section requires Section 2 for background. Specifically, Theorem 2.12
is required. If we were willing to accept that theorem without proof, the
required background would be reduced to Chapter I, the first eight sections
of Chapter II, and the first four sections of Chapter IV.
Given A = [a it ], B = (b lt . . . , bj, and C = [c x • • • cj, the standard
maximum linear programming problem is to find any or all nonnegative
X = (#!, . . . , x n ) which maximize
CX (3.1)
subject to the condition
AX < B. (3.2)
CX is called the objective function and the linear inequalities contained in
AX < B are called the linear constraints.
There are many practical problems which can be formulated in these
terms. For example, suppose that a manufacturing plant produces n different
kinds of products and that x, is the amount of they'th product that is produced.
Such an interpretation imposes the condition x s ;> 0. If c, is the income from
a unit amount of they'th product, then 2?=i c j x i = CX is the total income.
Assume that the objective is to operate this business in such a manner as to
maximize CX.
In this particular problem it is likely that each c, is positive and that CX
can be made large by making each x t large. However, there are usually
practical considerations which limit the quantities that can be produced.
For example, supppse that limited quantities of various raw materials to
make these products are available. Let b t be the amount of the ith ingredient
available. If a Xi is the amount of the rth ingredient consumed in producing
240 Selected Applications of Linear Algebra  VI
one unit of the y'th product, then we have the condition ]£j=i a u x i ^ ^
These constraints mean that the amount of each product produced must be
chosen carefully if CX is to be made as large as possible.
We cannot enter into a discussion of the many interesting and important
problems that can be formulated as linear programming problems. We
confine our attention to the theory of linear programming and practical
methods for finding solutions. Linear programming problems often involve
large numbers of variables and constraints and the importance of an efficient
method for obtaining a solution cannot be overemphasized. The simplex
method presented by G. B. Dantzig in 1949 was the first really prac
tical method given for solving such problems, and it provided the stimulus
for the development of an extensive theory of linear inequalities. It is
the computational method we describe here.
The simplex method is deceptively simple and it is possible to solve prob
lems of moderate complexity by hand with it. The rationale behind the
method is more subtle, however. We establish necessary and sufficient
conditions for the linear programming problem to have solutions and
determine procedures by which a proposed solution can be tested for opti
mality. We describe the simplex method and show why it works before giving
the details of the computational procedures.
We must first translate the statement of the linear programming problem
into the terminology and notation of vector spaces. Let U and V be vector
spaces over F of dimensions n and m, respectively. Let A = {<x 1? . . . , a n }
be a fixed basis of U and let B = {/3 l9 . . . , ft m } be a basis of V. If A = [a i:j ]
is a given m x n matrix, we let a be the linear transformation of U into V
represented by A with respect to A and B. Let P x be the finite cone in U
generated by A, and P 2 the finite cone in V generated by B.
If /? is the vector in V represented by B — (b x , . . . , b m ), the condition
A A
AX < B is equivalent to saying that <r(£) < /S. Let A be the basis in U
A A
dual to A and let B be the basis in V dual to B. Let <j> be the linear functional
A
in U represented by C = [c x , . . . c n ]. In these terms the standard maximum
linear programming problem is to find any or all I > which maximize
</>£ subject to the constraint tfd) < ft.
Let a be the dual of a. The standard dual linear programming problem
is to find any or all ip > which minimize tpfi subject to the constraint
a(xp) > <f). If we take a' = ' — a, ft' = — /?, and </>' = —<f>, then the dual
problem is to find a. ip > which maximizes xpfi' subject to the constraint
ff'CyO < 4> ■ Thus, the relation between the original problem, which we call
the primal problem, and the dual problem is symmetric. We could have
taken a minimum problem as the primal problem, in which case the dual
problem would have been a maximum problem. In this discussion, however,
we consistently take the primal problem to be a maximum problem.
3  Linear Programming 241
Any f > such that ct() < fi is called a feasible vector for the standard
primal linear programming problem. If a feasible vector exists, the primal
problem is said to be feasible. Any y > such that £ (ip) > <f> is called a
feasible vector for the dual problem, and if such a vector exists, the dual
problem is said to be feasible. A feasible vector £ such that <f>£ > 01 for
all feasible  is called an optimal vector for the primal problem.
Theorem 3.1. The standard linear programming problem has a solution
if and only if both the primal problem and the dual problem are feasible.
The dual problem has a solution if and only if the primal problem has a solution,
and the maximum value of '</>£ for the primal problem is equal to the minimum
value oj 'xpfi for the dual problem.
proof. If the primal linear programming problem is infeasible, then
certainly no optimum vector exists. If the primal problem is feasible, then
the assumption e a(P 1 ) + P 2 of Theorem 2.12 is satisfied. If the dual
problem is infeasible, then condition (2) of Theorem 2.12 cannot be satisfied.
Thus for every g there is a  > such that o(£) < /? and </>£ > g. This
means the values of <£! are unbounded and the primal problem has no
solution.
Now, assume that both the primal problem and dual problem are feasible.
If £ is feasible for the primal problem and \p is feasible for the dual problem,
then < xp{p  <?(£)} = #  y>o(g) = y>0  o(y>)£ = y>P  H + {</> ~
o(y>)} < # — <f>£. Thus cf>i is a lower bound for the values of y>8. Assume,
for now, that F is the field of real numbers and let g be the greatest lower
bound of the values of y>/ff for feasible y>. With this value of g condition (2)
of Theorem 2.12 cannot be satisfied, so that there exists a feasible  such that
<££o > g since <££o is also a lower bound for the values of y/S, </>!„ > g
is impossible. Thus <f>£ = g and <f>£ >: </>£ for all feasible f . Because of the
symmetry between the primal and dual problems the dual problem has a
solution under exactly the same conditions. Furthermore, since g is the
greatest lower bound for the values of %p{i, g is also the minimum value of y>0
for feasible ip.
If we permit F to be a subfield of the real numbers, but do not require
that it be the field of real numbers, then we cannot assert that the value of
g chosen as a greatest lower bound must be in F. Actually, it is true that g
is in F, and with a little more effort we could prove it at this point. However,
if A, B, and C have components in a subfield of the real numbers we can
consider them as representing linear transformations and vectors in spaces
over the real numbers. Under these conditions the argument given above
is valid. Later, when we describe the simplex method, we shall see that the
components of £ will be computed rationally in terms of the components
of A and B and will lie in any field containing the components of A, B, and C.
We then see that g is in F. □
242 Selected Applications of Linear Algebra  VI
Theorem 3.2. If £ is feasible for the standard primal problem and y> is
feasible for the dual problem, then £ is optimal for the primal problem and
xp is optimal for the dual problem if and only if y>{/3 — o"(£)} = and {d(ip) —
<£}£ = 0, or if and only if (f>tj = yfi.
proof. Suppose that £ is feasible for the primal problem and tp is feasible
for the dual problem. Then < \p{(5 — <7(£)} = ipfi — ya{£) = tpft —
#(yj)g = yp  <££ + {(f)  d(y>)}£ <ipfi <££• It is clear that y{p  <r(£)} =
and {a(tp) — (£}£ = if and only if ip(3 = <££.
If £ and tp are feasible and </>£ = y) P then <££ < ip f$ = <f>£o f° r a U
feasible £. Thus £ is optimal. A similar argument shows that ip Q is optimal.
On the other hand, suppose £ and ip are optimal. Let tp o p = g. Then
condition (2) of Theorem 2.12 cannot be satisfied for this choice of g. Thus,
there is a feasible £ such that <££ > g. Since £ is optimal, we have </>£„ >
</>£ > g = y) o p. Since t££ < ip @, this means </>£„ = y /5. n
Theorem 3.2 has an important interpretation in terms of the inequalities
of the linear programming problem as originally stated. Let £ = /S — <r(£)
be represented by Z = (z l9 . . . , z m ) and let r\ = <7(y>) — be represented
by W = [w 1  • • w n ]. Then the feasibility of £ and tp implies z t > and
w* > 0. The condition tpt, = 2f=i 2/j z ; = means t/ i z i = for each /.
Thus z i > implies y t = 0, and y t > implies z* = 0. This means that
if 2"=i a a x i ^ ^ * s satisfied as a strict inequality, then y t = 0, and if y t > 0,
this inequality must be satisfied as an equality. A similar relation holds
between the x j and the dual constraints. This gives an effective test for the
optimality of a proposed feasible pair X = (x u . . . , x n ) and Y = [y 1 • • • y m ].
It is more convenient to describe the simplex method in terms of solving
the equation cr(£) = (3 instead of the inequality <r(£) < /5. Although these
two problems are not equivalent, the two types of problems are equivalent.
In other words, to every problem involving an inequality there is an equiv
alent problem involving an equation, and to every problem involving an
equation there is an equivalent problem involving an inequality. To see
this, construct the vector space U ® V and define the mapping a x of U © V
into V by the rule cr 1 (£, rj) = <r(£) + r\. Then the equation o^g, rj) = (I
with £ > and r\ > is equivalent to the inequality cr(£) < /S with £ > 0.
This shows that to each problem involving an inequality there is an equivalent
problem involving an equation.
To see the converse, construct the vector space V © V and define the
mapping a 2 of U into V © V by the rule (7 2 (£) = (#(£), — ff(£)). Then the
inequality <7 2 (£) < (fj, — /S) with £ > is equivalent to the equation #(£) = ft
with £ > 0.
A.
Given a linear transformation a of U into V, ft e V, and <f> e U, the canonical
maximum linear programming problem is to find any or all £ > which
3  Linear Programming 243
maximize <j>£ subject to the constraint <r(f) = p. With this formulation
of the linear programming problem it is necessary to see what becomes of the
dual problem. Referring to a 2 above, for fa, yj e V ® V we have a z fa,
V 8 )£ = fa, V>J<*M = fa* Va)(<K£)> *(*» = Vi<*(£)  V2^(D = f (Vi 
V> a ). Thus, if we let ip = y> x — xp 2 , we see that we must have a fa =
# a (Vi» V») ^ <£ and (Vi» V2) e P 2~ © ^ But^the condition fa, ipj e
P% ® P£ is not a restriction on y> since any y> e V can be written in the form
^ = Vl — Va where v>i > ° and V2 ^ ° Thl i s ' the canonical dual linear
programming problem is to find any or all y e V which minimize yp subject
to the constraint a fa) > <f>.
It is readily apparent that condition (1) and (2) of Theorem 2.11 play
the same roles with respect to the canonical primal and dual problems that
the corresponding conditions (1) and (2) of Theorem 2.12 play with respect
to the standard primal and dual problems. Thus, theorems like Theorem
3.1 and 3.2 can be stated for the canonical problems.
The canonical primal problem is feasible if there is a I > such that
<r(f) = (} and the canonical dual problem is feasible if there is a y> e V such
that a fa) > <f>.
Theorem 3.3. The canonical linear programming problem has a solution
if and only if both the primal problem and the dual problem are feasible.
The dual problem has a solution if and only if the primal problem has a solution,
and the maximum value of '</>£ for the primal problem is equal to the minimum
value ofxpPfor the dual problem. □
Theorem 3.4. If I is feasible for the canonical primal problem and rp is
feasible for the dual problem, then I is optimal for the primal problem and y
is optimal for the dual problem if and only if {a fa — <f>}£ = 0, or if and only
From now on, assume that the canonical primal linear programming
problem is feasible; that is, P e o(PJ. There is no loss of generality in
assuming that a{Pj) spans V, for in any case p is in the subspace of V spanned
by o(PJ and we could restrict our attention to that subspace. Thus {afa),
. . . , o(<x n )} = o(A) also spans V and a(A) contains a basis of V. A feasible
vector £ is called a basic feasible vector if £ can be expressed as a linear
combination of m vectors in A which are mapped onto a basis of V. The
corresponding subset of A is said to be feasible.
Since A is a finite set there are only finitely many feasible subsets of A.
Suppose that £ is a basic feasible vector expressible in terms of the feasible
subset K, . . . , oc w }, I = 2*=i *i*< Then *(*) = 2S.1 ^W = £ Since
{<r(«i), . • • , tf(O) is a basis the representation of p in terms of that basis is
244 Selected Applications of Linear Algebra  VI
unique. Thus, to each feasible subset of A there is one and only one basic
feasible vector; that is, there are only finitely many basic feasible vectors.
Theorem 3.5. If the canonical primal linear programming problem is
feasible, then there exists a basic feasible vector. If the canonical primal
problem has a solution, then there exists a basic feasible vector which is
optimal.
proof. Let £ = ]££=! x i<*i De feasible. If {ofa), . . . , a(a k )} is linearly
independent, then £ is a basic feasible vector since {a l5 . . . , a J can be
extended to a feasible subset of A. Suppose this set is linearly dependent ;
that is, 2? =1 fiCfa) = where at least one t t > 0. Then JjLi ( x i — at i)
a(a f ) = j8 for every a e F. Let a be the minimum of xjti for those t t > 0.
For notational convenience, let x k /t k = a. Then x k — at k = and ^tli ( x i ~
at^ofai) = {}. If t i < we have x i — a?; > because a > 0. lft t > 0, then
x i — ati >x t — (xjt^tt = 0. Thus £' = 2*=i fat — ^O a i is a l so feasible
and expressible in terms of fewer elements of A. We can continue in this way
until we obtain a basic feasible vector.
Now, suppose the canonical problem has a solution and that £ as given
above is an optimal vector. If {a l5 . . . , a fc } is not a feasible subset of A, we
can assume x t > for i = I, . . . , k since otherwise £ could be expressed
in terms of a smaller subset of A. Let ip be an optimum vector for the dual
problem and let r\ = d(ip) — </> be represented by W = [vt^ • • • w n ]. By
Theorem 3.4 w t = for / = 1, . . . , k. It then follows that £' obtained as
above is also optimal. We can continue in this way until we obtain a basic
feasible vector which is optimal. □
If a linear programming problem has a solution, there remains the problem
of finding it. Since there are only finitely many basic feasible vectors
and at least one of them must be optimal, in concept we could try
them all and take the one that yields the largest value of </>£. It is not
easy, however, to find even one basic feasible vector, or even one feasible
vector, and for problems with large numbers of variables there would still
be an enormous number of vectors to test. It is convenient to divide the
process of finding a solution into two parts : to find a basic feasible vector
and to find an optimum vector when a basic feasible vector is known. We
take up the second of these problems first because it is easier to describe
the simplex method for it. It is then easy to modify the simplex method
to handle the first problem.
Now, suppose that 8 = {/? x , . . . , /5 TO } is a basis of V where & = ^(a;)
and {a 1? . . . , a TO } is a feasible subset of A. Let /? = 2^i b$i Then £ =
2^=i ^i a » ^ s t ne corresponding basic feasible vector and </>£ = 2*U C A
Suppose that a new coordinate system in V is chosen in which only one
basis element is replaced with a new one. Let /? r = a(a r ) be replaced by
3  Linear Programming 245
ft k = o(oL k ). Since a is represented by A = [a i} ] with respect to the bases
A and B we have ft k = 2™ i ^aA Since a rk #0 we can solve for ft r and
obtain
1 / m \
Pr = —\P*?.aM (33)
i ±r
Then
TO h W / tf \
P = I bA = P k + l(b i b r Aft,. (3.4)
i=i a rk i=i \ a rk J
If we let
a*
b k = ^ and b' i = b i b T ^ t (3.5)
u rk u rk
then another solution to the equation cr() = ft is
m
£' = &X + 1 ^. (3.6)
i=X
Notice that although ft remains fixed and only its coordinates change, each
particular choice of a basis leads to a different choice for the solution of the
equation cr(£) = ft.
We now wish to impose conditions so that £' will be feasible. Since b r >
we must have a rk > and, since b, > either a ik < or bja^ > b r \a rk .
This means r is an index for which a rk > and b r ja rk is the minimum of all
bja ik for which a ik > 0. For the moment, suppose this is the case. Then £'
is also a basic feasible vector.
Now,
TO
#' = c k b k + 2 c t b' t
i=\
h m / a
= cA + Ic i (b i b r ^
a rk i=x \ a rk
™ b m a
= 2 c i b i  c r b r + c k —  2 Cib r —
i=\ a rk i=i a rk
= <!>£ + — \c k  2 c i a ik)
a rk \ i=i J
= 0 + ^(c,4). (3.7)
a rk
Thus 4>g > 4>£ if c k  ^Zi c i a ik > °> and H' > H if also b r > 0.
246 Selected Applications of Linear Algebra  VI
The simplex method specifies the choice of the basis element /5 r = a(of. r )
to be removed and the vector fi k = o(a k ) to be inserted into the new basis
in the following manner:
(1) Compute 2™i cfiu = d, , for j = 1, . . . , n.
(2) Select an index k for which c k — d k > 0.
(3) For that k select an index r for which a rk > and b r \a rk is the minimum
of all bija ik for which a ik > 0.
(4) Replace (5 r = <r(a r ) by fc = a(a. k ).
(5) Express /? in terms of the new basis and determine f '.
(6) Determine the new matrix A' representing a.
There are two ways these replacement rules may fail to operate. There
may be no index k for which c k — d k > 0, and if there is such a k, there may
be no index r for which a rk > 0. Let us consider the second possibility
first. Suppose there is an index k for which c k — d k > and a ik < for
/ = 1, . . . , m. In other words, if this situation should occur, we choose to
ignore any other choice of the index k for which the selection rules would
operate. Then £ = a fc  J^ sl a ik ^ > and <r(£) = <r(a fc )  JZi «« ^fo) =
0. Also <j>^ = c k — 2™i c^,; > 0. Thus £ satisfies condition (1) of Theorem
2.14. Since condition (2) cannot then be satisfied, the dual problem is in
feasible and the problem has no solution.
Let us now assume the dual problem is feasible so that the selection pre
scribed in step (3) is always possible. With the new basic feasible vector
£' obtained the replacement rules can be applied again, and again. Since
there are only finitely many basic feasible vectors, this sequence of replace
ments must eventually terminate at a point where the selection prescribed
in step (2) is not possible, or a finite set of basic feasible vectors may be
obtained over and over without termination.
It was pointed out above that <£' > <f>£, and <(>£' > <f>£ if b r > 0. This
means that if b r > at any replacement, then we can never return to that
particular feasible vector. There are finitely many subspaces of V spanned
by m — 1 or fewer vectors in o"(A). If /3 lies in one of these subspaces, the
linear programming problem is said to be degenerate. If the problem is not
degenerate, then no basic feasible vector can be expressed in terms of m — 1
or fewer vectors in A. Under these conditions, for each basic feasible vector
every b t > 0. Thus, if the problem is not degenerate, infinite repetition is not
possible and the replacements must terminate.
Unfortunately, many practical problems are degenerate. A special
replacement procedure can be devised which makes infinite repetition
impossible. However, a large amount of experience indicates that it is very
difficult to devise a problem for which the replacement procedure given above
will not terminate.
3  Linear Programming 247
A.
Now, suppose Cj — dj < for j = 1 , . . . , n. Let 6 = {ip 1} . . . , y> m }
be the basis in V dual to 6, and consider ip = 2™i c iWi Then a(y>) =
L* i '<*(?<) = l£i *<(Z"i ««&} = 2"i (SZi ^ = 2F=x 4& > SLi
c^ = <£. Thus \p is feasible for the canonical dual linear programming
problem. But with this tp we have yj/5 = (2™ i c<Vi}{2j* i £*&} = 2™i C A
and ^ = {2?i^iKL=i^^} = 5ii c A This shows that # = <££•
Since both  and y> are feasible, this means that both are optimal. Thus
optimal solutions to both the primal and dual problems are obtained when
the replacement procedure prescribed by the simplex method terminates.
It is easy to formulate the steps of the simplex method into an effective
computational procedure. First, we shall establish formulas by which we
can compute the components of the new matrix A' representing a.
m
<*(*i) = 2 a ifii
m , to \
i=i a rk \ i=i /
= ^& + i(««'«*W (3.8)
a rk i=i\ a rk }
Thus,
a k j — , a tj — a^ a ik . \j")
a r1f a t
l rk u rk
It turns out to be more convenient to compute the new d'. — c 5 directly
Cj  d\ =  2 cfi't,  c k a' kj + C;
i=i
=  2 c i\ a u  — a ik)  c k — + C,
i=i \ a rk J a rk
= ( c ,rf,) — (c k d k ). (3.10)
a rk
For immediate comparison, we rewrite formulas (3.5) and (3.7),
K = ^, b' { = b,  ^ a ik , (3.5)
a rk a rk
<f>!' = <j>!; + ^{c k d k ). (3.7)
a T k
248
Selected Applications of Linear Algebra  VI
The similarity between formulas (3.5), (3.7), (3.9), and (3.10) suggests
that simple rules can be devised to include them as special cases. It is con
venient to write all the relevant numbers in an array of the following form :
c t
(3.11)
c k ~~ "k
c, — d,
The array within the rectangle is the augmented matrix of the system of
equations AX = B. The first column in front of the rectangle gives the
identity of the basis element in the feasible subset of A, and the second
column contains the corresponding values of c t . These are used to compute
the values of d^j = 1,...,«) and <f>£ below the rectangle. The row
[• • ■ c k • • • Cj •  •] is placed above the rectangle to facilitate computa
tion of the (c f — dj)(j = 1 , . . . , n). Since this top row does not change when
the basis is changed, it is usually placed at the top of a page of work for
reference and not carried along with the rest of the work. This array has
become known as a tableau.
The selection rules of the simplex method and the formulas (3.5), (3.7),
(3.9), and (3.10) can now be formalized as follows:
(1) Select an index k for which c k — d k > 0.
(2) For that k select an index r for which a rk > and b r \a rk is the minimum
of all bja^ for which a rk > 0.
(3) Divide row r within the rectangle by a rk , relabel this row "row k"
and replace c r by c k .
(4) Multiply the new row k by a ik and subtract the result from row /.
Similarly, multiply row k by (c fc — d k ) and subtract from the bottom row
outside the rectangle.
Similar rules do not apply to the row [• • • d k
dj • • •]. Once the
3  Linear Programming 249
bottom row has been computed this row can be omitted from subsequent
tableaux, or computed independently each time as a check on the accuracy
of the computation. The operation described in steps (3) and (4) is known
as the pivot operation, and the element a rk is called the pivot element. In the
tableau above the pivot element has been encircled, a practice that has been
found helpful in keeping the many steps of the pivot operation in order.
The elements c 3  — dj{j = 1 , . . . , ri) appearing in the last row are called
indicators. The simplex method terminates when all indicators are <0.
Suppose a solution has been obtained in which 8' = {fi[, . . . , ft^} is the
final basis of V obtained. Let 8' = {y)[, . . . , y)' m } be the corresponding dual
basis. As we have seen, an optimum vector for the dual problem is obtained
by setting y) = JjLi c«*)Y>i where i{k) is the index of the element of A mapped
onto fl' ki that is, <r(<x i(fc) ) — fi' k . By definition of the matrix A' = [a' i0 ] repre
senting a with respect to the bases A and 8', we have (3, = a(a.,) = 2k=i a i(k),jftk'
Thus, the elements in the first m columns of A' are the elements of the matrix
of transition from the basis 8' to the basis 8. This means that xp' k =
to m / m \
W=J, c i(k)W'k=1 c i(k)( 2X*uVj)
fc=l fc=l \j=l /
m i m \
= 2, I Z, C Hk) a i(k)J IVj
j=l\k=l ]
= Id' j y j . (3.12)
This means that D' = [d[ • ■ • d' m ] is the representation of the optimal
vector for the solution to the dual problem in the original coordinate system.
All this discussion has been based on the premise that we can start with
a known basic feasible vector. However, it is not a trivial matter to obtain
even one feasible vector for most problems. The z'th equation in AX = B is
n
^a ij x j = b f . (3.13)
i=i
Since we can multiply both sides of this equation by — 1 if necessary, there
is no loss of generality to assume that each b t > 0. We then replace equation
(3.13) by
SciijXj + Vi = b t . (3.14)
It is very easy to obtain a basic feasible solution of the corresponding system
of linear equations ; takea?! = ••■=#„ = OandUj = b { . We then construct
a new objective function
2c,*,M2> <f (3.15)
250
Selected Applications of Linear Algebra  VI
where M is taken as a number very much larger than any number to be
considered; that is, so large that this new objective function cannot be
maximized unless v y = • • • = v m = 0. The natural working of the simplex
method will soon bring this about if the original problem is feasible. At
this point the columns of the tableaux associated with the newly introduced
variables could be dropped from further consideration, since a basic feasible
solution to the original problem will be obtained at this point. However, it
is better to retain these columns since they will provide the matrix of tran
sition by which the coordinates of the optimum vector for the dual problem
can be computed. This provides the best check on the accuracy of the com
putation since the optimality of the proposed solution can be tested un
equivocally by using Theorem 3.4 or comparing the values of </>! and tp(3.
BIBLIOGRAPHICAL NOTES
Because of the importance of linear programming in economics and industrial engineering
books and articles on the subject are very numerous. Most are filled with bewildering,
tedious numerical calculations which add almost nothing to understanding and make
the subject look difficult. Linear programming is not difficult, but it is subtle and requires
clarity of exposition. D. Gale, The Theory of Linear Economic Models, is particularly
recommended for its clarity and interesting, though simple, examples.
EXERCISES
1. Formulate the standard primal and dual linear programming problems in
matric form.
2. Formulate the canonical primal and dual linear programming problems in
matric form.
3. Let A = [a^] be an m x n matrix. Let A 1X be the matrix formed from the
first r rows and first s columns of A ; let A 12 be the matrix formed from the first r
rows and last n — s columns of A ; let A 21 be the matrix formed from the last m — r
rows and first s columns of A; and let A 22 be the matrix formed from the last
m — r rows and last n — s columns of A. We then write
A =
A ti
We say that we have partioned A into the designated submatrices. Using this
notation, show that the matrix equation AX = B is equivalent to the matrix inequal
ity
" A~
X <
' B~
Y~ a \
\_BJ
4. Use Exercise 3 to show the equivalence of the standard primal linear pro
gramming problem and the canonical primal linear programming problem.
3  Linear Programming 251
5. Using the notation of Exercise 3, show that the dual canonical linear pro
gramming problem is to minimize
(Z l  Z 2 )B = [Z x Z 2 ]
B
subject to the constraint
[Zi Z 2 ]
[3*
and [Z 1 Z 2 ] > 0. Show that if we let Y = Z x — Z 2 , the dual canonical linear
programming problem is to minimize YB subject to the constraint YA ^ C without
the condition Y > 0.
6. Let A, B, and C be the matrices given in the standard maximum linear
programming problem. Let F be the smallest field containing all the elements
appearing in A, B, and C. Show that if the problem has an optimal solution, the
simplex method gives an optimal vector all of whose components are in F, and
that the maximum value of CX is in F.
7. How should the simplex method be modified to handle the canonical minimum
linear programming problem in the form: minimize CX subject to the constraints
AX = BandX>01
8. Find (x lt x 2 ) > which maximizes 5x x + 2x 2 subject to the conditions
2#i + x 2 < 6
4»! + x 2 <; 10
— x ± + x 2 < 3.
9. Find \y x y 2 t/ 3 ] > which minimize 6y x + \0y 2 + 3y 3 subject to the
conditions
Vi + Vi + Vs > 2
2«/i + Ay 2 y 3 >5.
(In this exercise take advantage of the fact that this is the problem dual to the
problem in the previous exercise, which has already been worked.)
10. Sometimes it is easier to apply the simplex method to the dual problem
than it is to apply the simplex method to the given primal problem. Solve the
problem in Exercise 9 by applying the simplex method to it directly. Use this work
to find a solution to the problem in Exercise 8.
11. Find (x x , x 2 ) > which maximizes x x + 2x 2 subject to the conditions
—x x + x 2 < 3
2x t + x 2 < 11.
12. Draw the lines
"~~ ZJJO* ~ f— **'2 ~ ~
2x x +a; 2 = ll
252 Selected Applications of Linear Algebra  VI
in the x x , ;z 2 plane. Notice that these lines are the extremes of the conditions given
in Exercise 11. Locate the set of points which satisfy all the inequalities of Exercise
11 and the condition (x x , x 2 ) > 0. The corresponding canonical problem involves
the linear conditions
— J*0Ct ~\~ X o "t~ X& — £
— 3?i "T" *^o ~ ' 4 ==
X* \ Xn "T" «^5 == /
2x x + x 2 + x 6 = 1 1 .
The first feasible solution is (0, 0, 2, 3, 7, 11) and this corresponds to the point
(0, 0) in the x x , x 2 plane. In solving the problem of Exercise 11, a sequence of
feasible solutions is obtained. Plot the corresponding points in the x lf x 2 plane.
13. Show that the geometric set of points satisfying all the linear constraints
of a standard linear programming is a convex set. Let this set be denoted by C.
A vector in C is called an extreme vector if it is not a convex linear combination
of other vectors in C. Show that if <f> is a linear functional and P is a convex linear
combination of the vectors {a 1? . . . , a r }, then <f>(p) < max {^(a,)}. Show that if p
is not an extreme vector in C, then either (/>(!) does not take on its maximum value
at p in C, or there are other vectors in C at which <£(£) takes on its maximum value.
Show that if C is closed and has only a finite number of extreme vectors, then the
maximum value of <£(^) occurs at an extreme vector of C.
14. Theorem 3.2 provides an easily applied test for optimality. Let X = (x lt . . . ,
x n ) be feasible for the standard primal problem and let Y = [y x  ■ ■ y m ] be feasible
for the standard dual problem. Show that X and Y are both optimal if and only
if x 3  > implies ^™ i Vi a a = c i and Vi > ° implies 2? =1 a ii x j = b t .
15. Consider the problem of maximizing 2x x + x 2 subject to the conditions
Xt ~~ £Xn ^^ J
x x < 9
2r x + x 2 < 20
x x + 3x 2 < 30
Consider X = (8, 4), Y = [0 1 0]. Test both for feasibility for the
primal and dual problems, and test for optimality.
16. Show how to use the simplex method to find a nonnegative solution of
AX — B. This is also equivalent to the problem of finding a feasible solution for
a canonical linear programming problem. (Hint: Take Z = (z x , . . . , z m ), F =
[1 1 • • • 1], and consider the problem of minimizing FZ subject to AX + Z = B.
What is the resulting necessary and sufficient condition for the existence of a solution
to the original problem?)
4  Applications to Communication Theory 253
17. Apply the simplex method to the problem of finding a nonnegative solution
of
6x 1 + 3x 2 — 4r 3 — 9x 4 — lx h — 5« 6 =
5^  8.r 2 + 8x 3 + 2x A  2x 5 + 5x 6 =
—1x x + 6x 2 — 5x 3 + 8a; 4 + 8.z 5 + x 6 =
x x + x 2 + x 3 + x i + x 5 + x 6 = 1.
This is equivalent to Exercise 5 of Section 1.
4 I Applications to Communication Theory
This section requires no more linear algebra than the concepts of a basis
and the change of basis. The material in the first four sections of Chapter I
and the first four sections of Chapter II is sufficient. It is also necessary that
the reader be familiar with the formal elementary properties of Fourier series.
Communication theory is concerned largely with signals which are uncer
tain, uncertain to be transmitted and uncertain to be received. Therefore,
a large part of the theory is based on probability theory. However, there are
some important concepts in the theory which are purely of a vector space
nature. One is the sampling theorem, which says that in a certain class of
signals a particular signal is completely determined by its values (samples)
at an equally spaced set of times extending forever.
Although it is usually not stated explicitly, the set of functions considered
as signals form a vector space over the real numbers; that is, if/(0 and g(t)
are signals, then (f + g)(t) = /(0 + g(t) is a signal and (af)(t) = af(t),
where a is a real number, is also a signal. Usually the vector space of signals
is infinite dimensional so that while many of the concepts and theorems
developed in this book apply, there are also many that do not. In many
cases the appropriate tool is the theory of Fourier integrals. In order to bring
the topic within the context of this book, we assume that the signals persist
for only a finite interval of time and that there is a bound for the highest
frequency that will be encountered. If the time interval is of length 1 , this
assumption has the implication that each signal /(/) can be represented as a
finite series of the form
N N
f(t) = £a + 2 <>* cos 27Tkt + 2 h sin 2irkt . (4. 1)
it=i fc=i
Formula (4.1) is in fact just a precise formulation of the vague statement
that the highest frequency to be encountered is bounded. Since the co
efficients can be taken to be arbitrary real numbers, the set of signals under
consideration forms a real vector space V of dimension 2N + 1. We show
that/(0 is determined by its values at IN + 1 points equally spaced in time.
This statement is known as the finite sampling theorem.
254 Selected Applications of Linear Algebra  VI
The classical infinite sampling theorem from communication theory
requires an assumption analogous to the assumption that the highest fre
quencies are bounded. Only the assumption that the signal persists for a finite
interval of time is relaxed. In any practical problem some bound can be
placed on the duration of the family of signals under consideration. Thus, the
restriction on the length of the time interval does not alter the significance
or spirit of the theorem in any way.
Consider the function
w(t) = ( 1 + 2 f cos 2nk t) e V. (4.2)
N
sin nt + ^ 2 cos 2nkt sin nt
y>(t) = *=1
(2N + 1) sin nt
N
sin nt + 2 sm (2nkt + nt) — sin {2nkt — irt)
(2N + 1) sin nt
= sin (2N + \)irt „ 3)
(2N + 1) sin nt
From (4.2) we see that ^(0) = 1 , and from (4.3) we see that y)(jj2N + 1) =
for < \j  < N.
Consider the functions
V*(0
= J t £ — \ for k = N,N + 1,...,N. (4.4)
V 2N + 1/
These 2N + 1 functions are all members of V. Furthermore, for tj =
jl(2N+ 1) we see that y>,(f,) = 1 while y> fc (^) = for k j±j. Thus, the
2N + 1 functions obtained are linearly independent. Since Vis of dimension
2N + 1, it follows that the set {ip k (t) \ k = — N, . . . , N} is a basis of V.
These functions are called the sampling functions.
If f(t) is any element of V it can be written in the form
fit) = I d k y> k {t). (4.5)
However,
f(h) = I d&A,) = d it (4.6)
or
/(0= I /(QV,(0 (47)
4  Applications to Communication Theory 255
Thus, the coordinates of /(f) with respect to the basis {y> k (t)} are (f(t_ N ),
. . . , f(t N )), and we see that these samples are sufficient to determine /(f).
It is of some interest to express the elements of the basis {\, cos 2irt, . . . ,
sin lirNt) in terms of the basis {ip k (t)}.
1 N
2 k=N
N
cos 2vjt = 2 cos 27T JhV>k(t) ( 4  8 )
iV
sin 2tt/7 = ^ sin 2ir/7 ifc v't(0
Jfc=iV
Expressing the elements of the basis (v^CO) i n terms of the basis {%, cos 27rf ,
. . . , sin lirNt) is but a matter of the definition of the y> k (t):
N l k \1
1 + 2Tcos2ir/t — )
2N + 1
= (l + 2 Y cos 2irjk cos 2irjt + 2V sin 2nJk sin 2tt/A
2iV+l\ i~i 2N + 1 sti 2N+1 J
(4.9)
With this interpretation, formula (4.1) is a representation of /(f) in one
coordinate system and (4.7) is a representation of /(f) in another. To
express the coefficients in (4.1) in terms of the coefficients in (4.7) is but a
change of coordinates. Thus, we have
"> = ^TTT ^ /(<*) C OS ^T = 7T^— 2/0*) COS 2TTJt k
2N + 1 *=# 2iV + 1 2AT + 1 k=N
2 » 2„ik 2 N (4  10)
b t = T f(t k ) sin Z7rJfc = 2 /(f fc ) sin 27TJt k .
\ 2N+U=iv 2W+1 2N+1*£n
There are several ways to look at formulas (4.10). Those familiar with
the theory of Fourier series will see the a, and bj as Fourier coefficients with
formulas (4.10) using finite sums instead of integrals. Those familiar
with probability theory will see the a, as covariance coefficients between the
samples of /(f) and the samples of cos 2njt at times t k . And we have just
viewed them as formulas for a change of coordinates.
If the time interval had been of length T instead of 1 , the series correspond
ing to (4.1) would be of the form
fit) = ia + 2>* cos £ t + £ b k sin ^ t. (4.11)
fc=l T k=l 1
256 Selected Applications of Linear Algebra  VI
The vector space would still be of dimension 2N + 1 and we would need
2N + 1 samples spread equally over an interval of length T, or (27V + l)/r
samples per unit time. Since N/T = W is the highest frequency present in
the series (4.11), we see that for large intervals of time approximately 2W
samples per unit time are required to determine the signal. The infinite
sampling theorem, referred to at the beginning of this section, says that if
W is the highest frequency present, then 2W samples per second suffice
to determine the signal. The spirit of the finite sampling theorem is in keeping
with the spirit of the infinite sampling theorem, but the finite sampling
theorem has the practical advantage of providing effective formulas for
determining the function f(t) and the Fourier coefficients from the samples.
BIBLIOGRAPHY NOTES
For a statement and proof of the infinite sampling theorem see P. M. Woodward,
Probability and Information Theory, with Applications to Radar.
EXERCISES
1. Show that y>(t r  t s ) = yl ^ M = <5 rs for N < r, s £ N.
2. Show that if /(f) can be represented in the form of (4.1), then
r l A
a k = 2 f{t) cos lnkt dt, k = 0, 1 , . . . , N,
C A
b k = 2 /(/) sin lirkt dt, k = 1, . . . , N.
r A i
3. Show that y(t) dt =
2N + 1
C A 1
4. Show that y k (t) dt =
JV2
in + r
5. Show that if /(f) can be represented in the form (4.7), then
£
l A N 1
This is a formula for expressing an integral as a finite sum. Such a formula is
called a mechanical quadrature. Such formulas are characteristic of the theory of
orthogonal functions.
6. Show that
N 1
2 OM _l_ 1 COS 27Trt * = a f*'
k=N zyv "+" l
4  Applications to Communication Theory 257
and
7. Show that
and
8. Show that
9. Show that
N l
k £ N 2N + l
N 1
2 i>r , 1 C0S 27rr ^ ~ '*) = S rk
fc=A 7 Liy + x
iV 1
2 ^m sin 27Tr C  '*) = °
k=N Zly + V
N
2 v*(0 = l.
fc=JV
fc=iV
10. If /(f) and^(f) are integrable over the interval [— , £], let
(/.*)= *f '/WO*.
J 14
Show that if/(0 and^(f) are elements of V, then {f,g) defines an inner product
in V. Show that I— = , cos 27rf, . . . , sin 2nNt\ is an orthonormal set.
11. Show that if /(f) can be represented in the form (4.1), then
f\4 a J N N
Show that this is Parseval's equation of Chapter V.
2
12. Show that
1
cos 2nrt wit) dt = ^^
13. Show that
r
rvi i
cos 27rr/ v»fc(0 dt = ^ r , t cos 2^.
J14
2N + 1
f 1 ^ 1
sin 2nrt y k {t) dt = ^^ , 1 sin 2irrt k .
14. Show that
r l A
I sin 7rrrt vu(t\ dt =
2N + 1
15. Show that
1
r (/)y,(f ) dt =
r l A
16. Using the inner product defined in Exercise 10 show that {y k (t)  k =
— TV, . . . , JV} is an orthonormal set.
1
258 Selected Applications of Linear Algebra  VI
17. Show that if /(f) can be represented in the form (4.7), then
Show that this is Parseval's equation of Chapter V.
18. Show that if /(f) can be represented in the form (4.7), then
/W<0* jjjfVt/CJ.
Show how the formulas in Exercises 12, 13, 14, and 15 can be treated as special
cases of this formula.
19. Let /(f) be any function integrable on [— , £]. Define
d k = (2N + \)( ' f(t) Vk (t)dt.
J\4
Then/ Ar (0 = ^i_ N d k y> k (t) e V. Show that
r l A rv*
/X0W(0 dt = f N it)tp r (t) dt, r=N,..., N.
J14 JH
Show that if g(t) e V, then
[* f{t)g{t)dt = [ f N {t)g(t)dt.
J l A JM
20. Show that if/ v (0 is defined as in Exercise 19, then
Show that
r'A r l A r l A
[/(» " fy(0f dt = /(f) 2 dt  f N {tf dt.
J l A JVz J\4
r l A ■ C l A
f N {tfdt<\ f{tfdt.
M J X A
21. Let^(f) be any function in V. Show that
f * [fit) giOfdt  [ ' [/(f) f x (t)fdt = [ A [f N (t) g(t)fdt.
J14 JV2 J\A
 X A
Show that
\ ' 1/(0 fsiOfdt < f 2 [/(f) g{t)fdt.
JV2 J l A
Because of this inequality we say that, of all functions in V,f^(0 is the best approxi
mation of /(f) in the mean; that is, it is the approximation which minimizes the
integrated square error. f N (t) is the function closest to/(f) in the metric defined by
the inner product of Exercise 10.
5  Vector Calculus 259
22. Again, let/(/) be a function integrable on [— \, \\ Define
Let c > be given. Show that there is an N(e) such that if N > N(e), then
r x A r l A r l A
f{t)dte<\ F N (t)dt<\ f(t)dt + e.
J14 J14 JVz
5 I Vector Calculus
We assume that the reader is familiar with elementary calculus, which
considers scalarvalued functions of a scalar variable and scalarvalued
functions of several scalar variables. Functions of several variables can also
be taken up from a vector point of view in which a set of scalar variables is
replaced by a vector. The background in linear algebra required is covered
in this text in the first three sections of Chapter IV and Sections 1 and 3 of
Chapter V.
In this section we consider vectorvalued functions of a vector variable.
We assume that the reader is acquainted in a different setting with most of
the topics mentioned in this section. We emphasize the algebraic aspects of
these topics and state or prove only those assertions of an analytic nature
that are intimately linked with the algebraic structure.
In this section we assume that V is a vector space of finite dimension n over
the real numbers or the complex numbers, and that a positive definite inner
product is defined in V. We shall write (a, /?) for the inner product of a,
ft g V. In Section 3 of Chapter V we showed that for any linear functional
<f> e V, there exists a unique r) e V such that <f>(p) = (r), (3) for all ft £ V. We
showed there that the mapping of <f> onto r)(<f>) = r\ defined in this way is
onetoone and onto.
We can use this mapping to define an inner product in V. Thus, we define
(<£, y) = (^), V (y)) = (itfvO, riffl). (5.1)
The conjugate appears in this definition because the mapping r\ is conjugate
linear and we require that the inner product be linear in the second variable.
It is not difficult to show that this does, in fact, define an inner product in V.
For the norm in V we have
UW* = (<£, +) = OM), n (4>)) = HOW. (5.2)
From Schwarz's inequality we obtain
IW)I = I0?(#, P)\ < I1 17(^)11 • WW = H\\ • \Wl (5.3)
260 Selected Applications of Linear Algebra  VI
Theorem 5.1. \\<f>\\ is the smallest value of M for which \<f>(p)\ < M \\p\\ for
all P e V.
proof. (5.3) shows that 0(0)1 < M \\p\\ holds for all if M = \\<f>\\. Let
P = titf). Then \cf>(P)\ = \(rj(<f>), fl\ = (P, P) = \\pV = H\\ ' WPW Thus,
the inequality \<j>{p)\ < M \\p\\ cannot hold for all values of p if M < \\<f>\\. U
Note. Although it was not pointed out explicitly, we have also shown that
for each <f> such a smallest value of M exists. When any value of M exists such
that \(f>(p)\ < M \\p\\ for all p, we say that (f> is bounded. Therefore, we have
shown that every linear functional is bounded. In infinite dimensional vector
spaces there may be linear functionals that are not bounded.
Iff is any function mapping U into V, we define
lim /() = a. (5.4)
to be equivalent to the following statement: "For any e > 0, there is a <5 >
such that HI —   < d implies /(!) — a < e." The function /is said
to be continuous at  if
lim /() = /( ). (5.5)
These definitions are the usual definitions from elementary calculus with
the interpretations of the words extended to the terminology of vector spaces.
These definitions could be given in other equivalent forms, but those given
will suffice for our purposes.
Theorem 5.2. Every (bounded) linear functional in V is continuous on all
ofV.
proof. Let M be any positive real number such that \<f>(P)\ < M \\p\\ holds
for all p e V. Then, for the given e > 0, it suffices (uniformly) to take d —
ejM. For any p o we have
W)  <t>{Po)\ = I W  AOI <M\\P P \\ < e (5.6)
whenever \\p — P \\ < <5. □
Theorem 5.3. Let A = {o^, . . . , oc n } be any basis in V. There exist positive
real numbers C and D, depending only on the inner product and the chosen
basis, such that for any  = 2j W =i x i*i G ^ we nave
n n
C2k<! <£>2K (5.7)
i=l
proof. By the triangle inequality
1111 =
2>* a i
<Ill^n =2teHki
5  Vector Calculus 261
Let D = max {aj}. Then
11*11 <l\Xi\' D = D l\*i\
On the other hand, let A = {(f> x , . . . , <f> n } be the dual basis of A. Then
W = \U0\ < \\<t>i\\ ' Ml Taking C 1 = 2Li HA > 0. we have
n n
2kl<2ll&IHI£ll =c~ 1 Mhn
i=l i=l
The interesting and significant thing about Theorem 5.3 is it implies that
even though the limit was defined in terms of a given norm the resulting
limit is independent of the particular norm used, provided it is derived from a
positive definite inner product. The inequalities in (5.7) say that a vector is
small if and only if its coordinates are small, and a bound on the size of the
vector is expressed in terms of the ordinary absolute values of the coordinates.
Let  be a vectorvalued function of a scalar variable /. We write  = £(f )
to indicate the dependence of  on t. A useful picture of this concept is to
think of  as a position vector, a vector with its tail at the origin and its head
locating a point or position in space. As t varies, the head and the point it
determines moves. The picture we wish to have in mind is that of £ tracing out
a curve as t varies over an interval (in the real case).
If the limit
lim ^ + h) ~ m  * (5.8)
ft>o h dt
exists, the vector valued function I is said to be differentiable at /. This
limit is usually called the derivative, but we wish to give this name to a diff
erent concept. At this moment the reasons for making the proposed dis
tinction would seem artificial and hard to explain. Therefore, we shall
reexamine this idea after we consider functions of vector variables.
,. !(* + h) 1(0 .
Since f (t + h) — (r) is a vector and h is a scalar, hm is a
vector. It is interpreted as a vector tangent to the curve at the point (f).
Now, let/ be a scalar valued function defined on V. Often, /is not defined
on all of V, but only on some subdomain. We do not wish to become in
volved in such questions. We assume that whenever we refer to the behavior
of/ at some £ e V,/() is also defined for all points in a sphere around £
of radius sufficiently generous to include all other vectors under discussion.
Let  be an arbitrary vector in V, which we take to be fixed for the moment.
Let r) be any other vector in V. For the given £ and r\, we assume the ex
pression m + *,)  /«.)
h
(5.9)
262 Selected Applications of Linear Algebra  VI
is defined for all h 5^ in an interval around 0. If
li„ /g« + *?>/<*■> ./■({.,,) (5.10)
h>o h
exists for each rj, the function /is said to be differentiable at  . It is con
tinuously dijferentiable at £ if/ is differentiable in a neighborhood around £„
and /'(, w) is continuous at  for each r\\ that is, lim/'(, rj) =/'( , ??).
We wish to show that/'(, 77) is a linear function of r\. However, in order
to do this it is necessary that/'(£, rj) satisfy some analytic conditions. The
following theorems lead to establishing conditions sufficient to make this
conclusion.
Theorem 5.4 {Mean value theorem). Assume thatf'(£ + hrj, rj) exists for all
h, < h < 1, and that /(£ + hrj) is continuous for all h, < h < 1. Then
there exists a real number 6, < Q < 1, such that
/(£ + V) /(£) =/"(£ + Ol* V) (511)
proof. Let g(h) =/( + to?) for f and ?? fixed. Then g(/i) is a real
valued function of a real variable and
g (ft) = hm —
Aft>0 Art
, . /(I + fo7+A/»y)/(g + f»7)
= hm ^ —
Afto Art
= /'(! + H»7) < 5 ' 12)
exists by assumption for < rt < 1. By the mean value theorem for g(h)
we have
g (l)g(0) = g'(6), O<0<1, (5.13)
or
/(I + n) /(« = /'(* + ^^ D
Theorem 5.5. Iff'(£, rj) exists, then for aeF. /'(£, 0*7) ex/ste and
fX£,ar,) = af'{S,rj). (5.14)
proof. / ( a^) = hm
^►o rt
,. f(£ + ahrj)m
= a hm
a/j>o art
= af'(5,rj)
for a 5* 0, and/'(f , fl»?) = for a = 0. □
5  Vector Calculus 263
Lemma 5.6. Assume f'(£ Q , Vi) exists and thatf'(£, tj 2 ) exists in a neighbor
hood of £„• (/"/'(£> V2) is continuous at £ , thenf'(^ , rj x + rj 2 ) exists and
/'do, Vi + %) = /'(&, Vi) + /'do, %)• (515)
PROOF.
/ do, % + Vt) = llm " 7
h*o n
.. /do + fah + fa?,)  fdo + H) + /do + fah) /do)
= 11m —  —
h*0 h
_ Bm /'(f. + hh + 9fch.feh) + nfi> %)
by Theorem 5.4,
= lim/'(lo + % + Ohrj 2 ,rj 2 ) + /'do »h)
by Theorem 5.5,
= /'do^ 2 )+/'do^i)
by continuity at £ . D
Theorem 5.7. Let A = {a 1? . . . , aj fee a fowis 0/ V over F. Assume
/'(lo, oO exists for all i, and thatf'(i, a*) exists in a neighborhood of £ and is
continuous at i for n — 1 of the elements of A. Thenf'(€ , rj) exists for all
rj and is linear in rj.
proof. Suppose /'(I, a^) exists in a neighborhood of  and is continuous
for 1 = 2, 3, . . . , n. Let S k = <a 1? . . . , a fc >. Theorem 5.5 says that/'d , rj)
is linear in rj for rj e S v
By induction, assume /'do, v) is linear in 77 for rj e S k . Then by Theorem
5.5, /'do, a k+1 ai k+1 ) exists in a neighborhood of £ c and is continuous at £
for all a fc+1 e F. By Lemma 5.6, /'d , V + 0*+i a *+i) = /'do, *?) + /'do»
«*+i a *+i) =/'do, »?) + 0*+i/'do, «wi). Since all vectors in S k+1 are of the
form rj + tf fc+ ia fc+1 , /'do, ??) is linear for all rj e S k+1 . Finally, /'d , rj) is
linear for all rj eS n = V.o
Theorem 5.7 is usually applied under the assumption that/'d, rj) is con
tinuously differentiable in a neighborhood of  for all r) eV. This is certainly
true if/'d, 0O is continuously differentiable in a neighborhood of £ for all 0^
in a basis of V. Under these conditions, /'(f , 77) is a scalar valued linear
function of rj denned on V. The linear functional thus determined depends
on £ and we denote it by df(!j). Thus df(t) is a linear functional such that
df(£)(rj) =/'d, rj) for all jjeV. #(£) is called the differential off at f.
264 Selected Applications of Linear Algebra  VI
If A = {a 1? . . . , aj is any basis of V, any vector £ eV can be represented
in the form £ = 2* x i 0L i Thus, any function of  also depends on the
coordinates (z ± , . . . , x n ). To avoid introducing a new symbol for this
function, we write
/(I) =/(*i, •••,*«)• (516)
Since df(tj) is a linear functional it can be expressed in terms of the dual
basis A = {&,... , <£ J. The coordinates of df(fj) are easily computed by
evaluating df(£) for the basis elements oq e A. We see that
df(£K*i) = / (5. a<) = lim
= lim
h
f(x lt . . . , x t + h, . . . , x n )  f(x lt ...,x n )
h
(5.17)
Thus,
dm = lf^ (518)
a
'#„•
For any
i
df(m=f'(t,v) = I§ L yi (5i9)
From (5.17), the assumption that /'(!, ??) is a continuous function of £
implies that the partial derivatives are continuous. Conversely, the con
tinuity of the partial derivatives implies that the conditions of Theorem 5.7
are satisfied and, therefore, that /'(£, rj) is a continuous function of f. In
either case, /'(I, rj) is a linear function of rj and formula (5.19) holds.
Theorem 5.8. Iff'(£, rj) is a continuous function of  for all rj, then df{£)
is a continuous function of £.
proof. By formula (5.17), if /'(I, rj) is a continuous function of , then
J = f (£, a) is a continuous function of f . Because of formula (5.18) it
a^ 7 v * y
then follows that df{£) is a continuous function of £. D
If ^ is a vector of unit length, /'(I, rj) is called the directional derivative of
/() in the direction of rj.
Since
we see that
5  Vector Calculus 265
Consider the function that has the value x t for each  = 2* x^. Denote
this function by X t . Then
dXlgfaj) = hm f
ft»o n
— d a
dxm = &.
It is more suggestive of traditional calculus to let x t denote the function X { ,
and to denote ^ by dx t . Then formula (5.18) takes the form
df=2^dx i . (5.18)
i OX i
Let us turn our attention for a moment to vectorvalued functions of vector
variables. Let U and V be vector spaces over the same field (either real or
complex). Let U be of dimension n and V of dimension m. We assume that
positive definite inner products are defined in both U and V. Let F be a
function defined on U with values in V. For £ and r\ e U, F'(£, rj) is defined
by the limit, ftm , N w . x
ftt,,)lim ftf + * , '> fffl , (5.20)
if this limit exists. If F'(£, rj) exists, F is said to be differ entiable at . F is
continuously differentiable at  if Fis differentiable in a neighborhood around
 and F'(£, rj) is continuous at £ for each rj; that is, HmF'(£, 17) =
F'(£ , *?)• *^°
In analogy to the derivative of a scalar valued function of a vector variable,
we wish to show that under appropriate conditions F'(£, rj) is linear in rj.
Theorem 5.9. If for each rj e U, F'(£, rj) is defined in a neighborhood of 
and continuous at £ , then F'(£o> V) /5 linear in rj.
proof. Let ip be a linear functional in V. Let/be defined by the equation
/(£) = y{F()}. (5.21)
Since ip is linear and continuous,
_ lim jm+JstLium
h*0 [ h )
\h^0 h J
= W {F'{i,rj)}. (5.22)
266 Selected Applications of Linear Algebra  VI
Since ip is continuous and denned for all of V,/'(f , rj) is defined in a neighbor
hood of £ and continuous at  . By Theorem 5.7, /'( , rj) is linear in rj.
Thus,
y>{F'(g , a 1 r] 1 + a 2 rj 2 )} = a x %p{F'(£ Q , ^)} + a 2 y){F'(£ , rj 2 )}
= tpfaF'iZo, rjj + flaF'Clo. %)}•
Since F'(£ , «i»h + «2%)  «i*"(£o» »7i)  «2*"(? > %) is annihilated by all
A.
^> e y, it must be 0; that is,
F'($o, aiVx + «2^ 2 ) = a.F'i^o, ^) + a 2 F'( , r) 2 ). D (5.23)
For each , the mapping of r\ e U onto F'(£, ??) e Vis a linear transforma
tion which we denote by F'(£). F'() is called the derivative of Fat £.
It is of some interest to introduce bases in U and V and find the correspond
ing matrix representation of F'(g). Let A = {a lf . . . , a B } be a basis in U
and 8 = {&, ... , pj be a basis in V. Let £ = ]£, x^ and F(£) = J, */,•(£)&•
Let B = (vi, • • • , ^ TO } be the dual bais of B  Then VfcW) = &(£)• If
F'(£) is represented by the matrix / = [a i} ], we have
F'M*,) = I *iA ( 5  24 )
Then
r F(g + h* t )  F(l)
y> fc F(£ + ftoc ,)  y k F(£)
= hm
= lim y ^ + h(Xj) ~ Vk ^
h>0 h
dx j
according to formula (5.17). Thus F'(£) is represented by the matrix
(5.25)
J(S) =
' dy k
dX;
(5.26)
7(1) is known as the Jacobian matrix of F'(£) with respect to the basis A and 8.
The case where U = V — that is, where Fis a mapping of V into itself— is of
special interest. Then F'(£) is a linear transformation of V into itself and the
corresponding Jacobian matrix 7() is a square matrix. Since the trace of J()
is invariant under similarity transformation, it depends on F'(£) alone and not
5  Vector Calculus 267
on the matrix representation of F'(£). This trace is called the divergence
of F at £.
Tr(F'(l)) = divF(£) = i^. (5.27)
1=1(7^
Let us reexamine the differentiation of functions of scalar variables,
which at this point seem to be treated in a way essentially different from
functions of vector variables. It would also be desirable if the treatment of
such functions could be made to appear as a special case of functions of
vector variables without distorting elementary calculus to fit the generaliza
tion.
Let W be a 1dimensional vector space and let {yj be a basis of W. We
can identify the scalar t with the vector ty x , and consider (f) as just a short
handed way of writing (fy x ). In keeping with formula (5.10), we consider
Um ««?, + hi)  Htn) _ {Xtyu v) (5 28)
h>0 h
Since W is 1dimensional it will be sufficient to take the case r\ = y x . Then
ZVyiXri) = ZVyi, ?i) = hm 1
h>o n
= lim «' + h) ~ «'>
7i»0 h
= *i . (5.29)
dt
Thus dgjdt is the value of the linear transformation l'(/yi) applied to the
basis vector y x .
Theorem 5.10. Let F be a mapping of U into V. If F is linear, F'(£) = F
for all feU.
proof. F (!)w = F (I, ?y) = hm
a>o n
= lim F(^)
= F(rj). □ (5.30)
Finally, let us consider the differentiation of composite functions. Let F
be a mapping of U into V, and G a mapping of V into W. Then GF = H
is a mapping of U into W.
Theorem 5.11. If F is linear and G is differentiate, then (GF)'Q) =
G'(F(£))F. IfG is linear and F is differentiate, then (GF)'() = GF'(£).
268 Selected Applications of Linear Algebra  VI
proof. Assume F is linear and G is differentiable. Then
GF(£ + hri)  GF(£)
(GF)'(l)0?) = (GF)X£,V) = Mm
ft»o h
G[F() + hF(rj)]  G[F(m
= lim
h*0 h
= G'[F(i),F(rj)]
= G'(F(*))Ffo).
Assume G is linear and F is differentiable. Then
(GF)'(0(V) = W(f, 17) = lim GF( * + ^
(5.31)
GFtf)
— Ill
h>
ii
h
= lim G
ft»0
rF(i + ^) 
 F{&]
h
_
= G
lim
_h>0
F( + fa?) 
 F(m
h
_
= G[F'(£,rj)]
= G[F'(0fo)]
= [GFXmv) □
(5.32)
Theorem 5.12. Let F be a mapping of U into V, and G a mapping of V into
W. Assume that Fis continuously differentiable at £ c and that G is continuously
differentiable in a neighborhood of t = F(£ ). Then GF is continuously
differentiable at  and (GF)'( ) = G'(F(£ ))F'(g ).
proof. For notational convenience let F( + hr\) — F(f ) = co. Let y>
be any linear functional in W. Then ipG is a scalarvalued function on U
continuously differentiable in a neighborhood of t . Hence
y>GF(f + hr[)  y>GF(£ ) = V>G(t + co)  y>G(r )
= (vG)'(t + 6a),co)
for some real d, < 6 < 1. Now
F(l + foj)  F( )
lim — = lim
7»»0 /t
and
ft+o n
= F'{^,ri)
lim (t + 0co) = r .
5  Vector Calculus 269
Since G'(r, co) is continuous in t in a neighborhood of t and bounded linear
in ft) ' v<Gm$M) = (wGF)'(h, n)
— lim  (y>G)'(t + Oco, oS)
ft»0 /i
= lim \pG' j t + 6oj, — \
a>o \ h )
= y)G'(r , F(£o, ??))
= yG'(F(fo)>n£o)fo))
= ^[G'(nio))F'(^o)](^). (533)
Since (5.33) holds for all y> e W, we have
(GF)'(lo)(^) = G'(F( ))F'(lo)(^). □ (534)
This gives a very reasonable interpretation of the chain rule for differentia
tion. The derivative of the composite function GF is the linear function
obtained by taking the composite function of the derivatives of F and G.
Notice, also, that by combining Theorem 5.10 with Theorem 5.12 we can
see that Theorem 5.11 is a special case of Theorem 5.12. If F is linear
(GF)'(i) = G'(F(£))F'(£) = G'(F(0)F If G is linear, (GF)'() = G'(F(£))F'
(£) = GF'(I).
Example 1. For f = xj x + z 2 £ 2 , let/(f) = x* + 3x 2 2  2x x x 2 . Then for
V = 2/ifi + VoJ 2 we have
/ (, ^) = hm •>
&>o n
= 2^^ + 6x 2 ?/ 2 — 2x x y 2 — 2x 2 y x
= (2x 1 — 2x 2 )y 1 + (6x 2 — 2x x )y 2 .
A
If {^> l5 <f> 2 } is the basis of R 2 dual to {£ l5 £ 2 }, then
/'(flfo) = /'(£ i?)
= [(2^  2^)0! + (6x 2  2x 1 )<f> 1 ](r ] ),
and hence
/'(I) = (2x x  2s 2 )& + (6* 2  2^)^
dL , , 9L .
d^ cto 2
Example 2. For  = a^lx + z 2  2 + x^ z , letF() = sin x z £ x + a^a^la +
(V + aj 3 *) 8 . Then for rj = y^ + y 2 £ 2 + 2/ 2 £ 3 we have
F'(£, fj) = y 2 cos ^Ix + {x 2 x z y x + ar^gi/a + x x x 2 y 3 )^ 2 + (2x 1 y 1 + 2a: 3 y 3 )3
= F'(0(V)
270 Selected Applications of Linear Algebra  VI
We see that F'(g) is a linear transformation of R 3 into itself represented with
respect for the basis {f 1;  2 , f 3 } by the matrix
cos x 2
**/rt»^Q %Ast%Aj*y lAyl^VO
In both of these examples rather conventional notation has been used.
However, the interpretation of the context of these computations is quite
different from the conventional interpretation. In Example 1 , /(I) should
be regarded as a function mapping R 2 into R, and/'(£) should be regarded
as a linear approximation of/at the point £. Thus/'(£) g Horn (R 2 , R) = R 2 .
In Example 2, F(£) is a function of R 3 into R 3 and F'(£) is a linear approxi
mation of Fat the point f. Thus F'(£) e Horn (R 3 , R 3 ).
6 I Spectral Decomposition of Linear Transformations
Most of this section requires no more than the material through Section 7
of Chapter III. However, familiarity with the Jordan normal form as
developed in Section 8 of Chapter III is required for the last part of this
section and very helpful for the first part.
Let a be a linear transformation of an wdimensional vector space V into
itself. The set of eigenvalues of a is called the spectrum of a. Assume that
V has a basis {a 1} . . . , a n } of eigenvectors of a. Let \ be the eigenvalue
corresponding to a^. Let S { be the subspace spanned by a*, and let 7^ be the
projection of V onto S t along S x ©
projections have the properties
and
TTfTj = for
© s,_ x © s i+1 © •
■®s n .
These
(6.1)
i*j.
(6.2)
Any linear transformation a for which a 2 = a is said to be idempotent.
If a and r are two linear transformations such that ot = ra = 0, we say
they are orthogonal. Similar terminology is applied to matrices representing
linear transformations with these properties.
Every  e V can be written in the form
£ = fi + ■ • ■ + £„. (6.3)
where £ t e S^. Then tt^I) = £ f so that
I = ir x (£) + • • • + 7r„(£)
= (tt! + • • • + ttJ(I). (6.4)
6  Spectral Decomposition of Linear Transformations 271
Since (6.4) holds for every £ e V we have
1 = tt, + 7T 2 + ' ' ' + TT n  ( 6 5)
A formula like (6.5) in which the identity transformation is expressed as
a sum of mutually orthogonal projections is called a resolution of the identity.
From (6.1) and (6.2) it follows that (ttj + 7r 2 ) 2 = K + tt 2 ) so that a sum of
projections orthogonal to each other is a projection. Conversely, it is some
times possible to express a projection as a sum of projections. If a projection
cannot be expressed as a sum of nonzero projections, it is said to be irreduc
ible. Since the projections given are onto 1dimensional subspaces, they are
irreducible. If the projections appearing in a resolution of the identity are
irreducible, the resolution is called irreducible or maximal.
n
Now, for £ = 2 li as in ( 6  3 ) we have
\ 1=1 / i=l
n
= 1 ^£i
1=1
n
= I WS)
(0. (66)
Since (6.6) holds for every  £ V we have
(7 = 2^. (6.7)
= ( 2 V<)<
i=l
A representation of cr in the form of (6.7), where each A* is an eigenvalue of
a and each 7T t is a projection, is called a spectral decomposition. If the
eigenvalues are each of multiplicity 1, the decomposition is unique. If
some of the eigenvalues are of higher multiplicity, the choice of projections
is not unique, but the number of times each eigenvalue occurs in the decom
position is equal to its multiplicity and therefore unique.
An advantage of a spectral decomposition like (6.7) is that because of
(6.1) and (6.2) we have
cr 2 = iV^, (6.8)
i=l
tf* = JU**<, (69)
272 Selected Applications of Linear Algebra  VI
and
/(*) = 1KK>"< (610)
for any polynomial/^) with coefficients in F.
Given a matrix representing a linear transformation a, there are several
effective methods for finding the matrices representing the projections 7r i5
and, from them, the spectral decomposition. Any computational procedure
which yields the eigenvectors of a must necessarily give the projections since
with the eigenvectors as basis the projections have very simple representations.
However, what is usually wanted is the representations of the projections in
the original coordinate system. Let S, = (s 1:j , s 2j , . . . , s ni ) be the repre
sentation of a, in the original coordinate system. Since we have assumed there
is a basis of eigenvectors, the matrix S = [.%] is nonsingular. Let T = [t i:j ]
be the inverse of S. Then
• P k = [Silch,] (6.11)
represents 7r fc , as can easily be checked.
We give another method for finding the projections which does not require
finding the eigenvectors first, although they will appear in the end result. We
introduce this method because it is useful in situations where a basis of
eigenvectors does not exist. We, however, assume that the characteristic
polynomial factors into linear factors. Let {A l5 . . . , X p ) be the distinct
eigenvalues of a, and let
m(x) = (x  A x ) s i ■••(* X v ) s » (6.12)
be the minimum polynomial for a. Set
K(x) = 7 Jn ^~ s .. (6.13)
We wish to show now that there exist polynomials giCe), • • • , g p ( x ) such that
1 = £i(*)fti(*) + • • • + g 9 {x)h p (x). (6.14)
Consider the set of all possible nonzero polynomials that can be written
in the form /? i (a;)A 1 (a;) + • • • + p p {x)h v {x) where the Pi(x) are polynomials
(not all zero since the resulting expression must be nonzero). At least one
polynomial, for example, h^x), can be written in this form. Hence, there is a
nonzero polynomial of lowest degree that can be written in this form. Let
d(x) be such a polynomial and let g^x), . . . , g p (x) be the corresponding
coefficient polynomials,
d(x) = g 1 (x)h 1 (x) + • • • + g p (x)h p (x). (6.15)
6  Spectral Decomposition of Linear Transformations 273
We assert that d(x) divides all h t {x). For example, let us try to divide h x {x)
by d(x). Either d(x) divides h x {x) exactly or there is a remainder r x (x) of
degree less than the degree of d(x). Thus,
h x {x) = d(x) qi (x) + ri (x) (6.16)
where q x {x) is the quotient. Suppose, for the moment, that the remainder
r x {x) is not zero. Then
r x {x) = h x (x)  d(x)q 1 (x)
= Ai(a;){l  giO)?iO)}  g2(x)qi(x)h 2 (x) g p {x) qi {x)h v {x).
(6.17)
But this contradicts the selection of d(x) as a nonzero polynomial of smallest
degree which can be written in this form. Thus d(x) must divide h^x).
Similarly, d(x) divides each h^x).
Since the factorization of m(x) is unique, the h t {x) have no common non
constant factor and d{x) must be a constant. Since we can divide any of
these expressions by a nonzero constant without altering its form, we can
take d{x) to be 1. Thus, we have an expression in the form of (6.14).
If we divide (6.14) by m(x) we obtain
1 gl(x) +•••+ 8 » (X) . (6.18)
m{x) {x  A x ) Sl (x  XJ*
This is the familiar partial fractions decomposition and the polynomials
gi(x) can be found by any of several effective techniques.
Now setting e t {x) = g^h^x), we see that we have obtained a set of
polynomials {e x {x), . . . , e v {x)} such that
(1) 1 = e x {x) + ••• + e v (x),
(2) e i (x)e j (x) is divisible by m(x) for i ^ j.
(3) (x — l^) Si e^x) is divisible by m(x).
Now, we use these polynomials to form polynomial expressions of the
linear transformation a. Then {e^a), . . . , e p (a)} is a set of linear trans
formations with the properties that
(1) 1 = ei (a) + ■ • • + e p (a),
(2) e i (a)e j (o) = 0fori^j,
(3) (a  We^o) = 0. (6.19)
From (1) and (2) it follows, also, that
(4) e t (&) = 1 • e t {a) = (e^a) + • • • + e v (o)) ei (a)
= e^e^a) H h e v (a)e t (a)
274 Selected Applications of Linear Algebra  VI
These four properties suffice to show that the e^a) are mutually orthogonal
projections. From (3) we see that e<(o)(V) is in the kernel of (<r — A f ) s \ As in
Chapter III8, we denote the kernel of (a  A^) 8 * by M t . Since e t (x) is
divisible by (x  A,) 8 ' fory 5* i, e^a)^ = 0. Hence, if ft e M„ we have
Pi = (*i(*) + • • • + *,(<r))(ft)
= *<(*)(&)• (6.20)
This shows that e t (&) acts like the identity on M t . Then M t = e^aXM^ c
ei(a)(V) <= M. so that e,(<r)(V) = M 4 . By (6.19), for any £ e V we have
= (e,(cr) + • • • + e»)(ft
= h + • • • + j8 p , (6.21)
where ft = ^(cr)(ft e M,. If £ = 0, then ft = ^(cr)(ft = ^(ct)(0) =
so that the representation of in the form of a sum like (6.21) with ft e M { is
unique. Thus,
V = M . e • • • © M p . (6.22)
This provides an independent proof of Theorem 8.3 of Chapter III.
Let (?j = a  e t (a). Then
(1) a = a x + • • • + o p ,
(2) tfiOj = for i 5^ j, and
(3)f(a)=f(a 1 ) + +f(a p ), (6.23)
for any polynomial/^) with coefficients in F. If^ = 1, then (c; — h^e^a) =
so that a t = ^e^a). In this case M i is the eigenspace corresponding to A*.
If the multiplicity of X t is 1, then dim M t = 1 and e^a) = ir i is the projection
onto M^. If A represents a, then P, = e t (A) represents tt> If the multiplicity
of Aj is greater than 1, then e t (a) is a reducible projection. e t (cr) can be
reduced to a sum of irreducible projections in many ways and there is no
reasonable way to select a unique reduction.
If Si > 1 , the situation is somewhat more complicated. Let r i = (a —
X^e^a). Then v s J = (a — X^e^a) = 0. A linear transformation for which
some power vanishes is said to be nilpotent. Thus,
a { = A^(<7) + Vi (6.24)
is the sum of a scalar times an indempotent transformation and a nilpotent
transformation.
Since a commutes with e t (a), a(Mi) = <r^(<7)(V) = e^aiV) c M*. Thus
M i is invariant under a. It is also true that a^M^ <= M t and ff^M,) = {0}
for j ^ i. Each M, is associated with one of the eigenvalues. It is often
possible to reduce each M t to a direct sum of subspaces such that a is invariant
on each summand. In a manner analogous to what has been done so far,
6  Spectral Decomposition of Linear Transformations 275
we can find for each summand a linear transformation which has the same
effect as a on that summand and vanishes on all other summands. Then this
linear transformation can also be expressed as the sum of a scalar times an
idempotent transformation and a nilpotent transformation. The deter
mination of the Jordan normal form involves this kind of reduction in some
form or other. The matrix B t of Theorem 8.5 in Chapter III can be written
in the form A*E, + N> where E t is an idempotent matrix and N t is a nilpotent
matrix. However, in the following discussion (6.23) and (6.24) are sufficient
for our purposes and we shall not concern ourselves with the details of a
further reduction at this point.
There is a particularly interesting and important application of the spectral
decomposition and the decomposition (6.23) to systems of linear differential
equations. In order to prepare for this application we have to discuss the
meaning of infinite series and sequences of matrices and apply these decom
positions to the simplification of these series.
If {A k }, A k = [a w (fc)], is a sequence of real matrices, we say
lim^ fc = A = [a tj ]
k
if and only if lim a^k) = a {j for each / and/ Similarly, the series ]£* =0 c k A k
is said to converge if and only if the sequence of partial sums converges.
It is not difficult to show that the series
yA k (6.25)
*ofc!
converges for each A. In analogy with the series representation of e x , the
series (6.25) is taken to be the definition of e A .
If A and B commute, then (A + B) k can be expanded by the binomial
theorem. It can then be shown that
e ^+B _ g A gB
This "law of exponents" is not generally satisfied if A and B do not commute.
If E is an idempotent matrix — that is, E 2 = E — then
k\
= I  E + E(l + X + • • • + ^ +   \
= i + («,* _ i)E. (6.26)
If TV is a nilpotent matrix, then the series representation of e^ terminates
after a finite number of terms, that is, e" is a polynomial in N. These
observations, together with the spectral decomposition and the decomposition
276
Selected Applications of Linear Algebra  VI
(6.23) in terms of commuting transformations (and hence matrices) of these
types, will enable us to express e A in terms of finite sums and products of
matrices.
Let a = a x + • • • + a v as in (6.23). Let A represent a and A t represent
Gi. Let E t represent e^a) and N t represent v t . Since each of these linear
transformations is a polynomial in a, they (and the matrices that represent
them) commute. Assume at first that a has a spectral decomposition, that
is, each v t = 0. Then A t = A^ and
gA _ g^iH VA V
— e Ai e A 2 . . . e
= [I + (e^  1)£J[/ + (**■  l)E 2 ] •••[/ + (e x »  1)E P ]
= / + £(«*' 1)E, (6.27)
because of the orthogonality of the E t . Then
e A = 2 e A % (6.28)
because ^? =1 Et = I. This is a generalization of formula (6.10) to a function
of A which is not a polynomial.
The situation when a does not have a spectral decomposition is slightly
more complicated. Then A { = A^ + N t and
Ni
e A = fle XEi fle
As an example of formula (6.28) consider A =
2 1
(6.29)
The character
istic polynomial is x* — x — 6 = (x — 3) (x + 2). To fix our notation we
take A x = — 2 and A 2 = 3. Since
1
+
(# — 3)(# + 2) x + 2 x — 3
 (a;  3) (a; + 2)
we see that e x (x) = and e 2 (a;) = • Thus,
F — — i
^1 — 5
£* =
1 2"
2 4_
/ = J&! + £ 2 ,
A = —2E X + 3is 2 ,
e A = e ~*E x + e 3 £ 2 .
'4 2"
2 1
6  Spectral Decomposition of Linear Transformations
277
BIBLIOGRAPHICAL NOTES
In finite dimensional vector spaces many things that the spectrum of a transformation
can be used for can be handled in other ways. In infinite dimensional vector spaces matrices
are not available and the spectrum plays a more central role. The treatment in P. Halmos,
Introduction to Hilbert Space, is recommended because of the clarity with which the
spectrum is developed.
EXERCISES
1 . Use the method of spectral decomposition to resolve each matrix A into the
form A = AE X + AE 2 = X X E X + X 2 E 2 , where X x and X 2 are the eigenvalues of A
and E x and E 2 represent the projections of V onto the invariant subspaces of A.
(a)
"1 2"
(b)
"1
2"
(c)
"1 2"
2 1_
»
.2 2_
»
_2 4_
id)
"1 3"
to
"1 4"
3 
7_
5
4 
5_
2. Use the method of spectral decomposition to resolve the matrix
"3 7 20"
A =
■14
5
3
into the form A = AE X + AE 2 + AE 3 = X X E X + h 2 E 2 + ^E 3 , where E lf E 2 , and
E 3 are orthogonal idempotent matrices.
3. For each matrix A in the above exercises compute the matrix form of e A .
4. Use the method of spectral decomposition to resolve the matrix
13 3"
32
1 1 1
A =
into the form A = AE X + AE 2 where E x and E 2 are orthogonal idempotent
matrices. Furthermore, show that AE ± (or AE 2 ) is of the form AE X = XJE^ + A^
where X x is an eigenvalue of A and N x is nilpotent, and that AE 2 (or AE X ) is of the
form AE 2 = X 2 E 2 where A 2 is an eigenvalue of A.
5. Let A = XE + N, where E 2 = E, EN = N, and JV is nilpotent of order r;
that is, N 1  1 7* but N r = 0. Show that
rl fts
e A = I + E(e x  1) + e* T —
«ti si
= (/  E){\  e x ) + e k e N =1 E + e x e N E.
278 Selected Applications of Linear Algebra  VI
6. Compute the matrix form of e A for the matrix A given in Exercise 4.
7. Let A = AE X + AE 2 + ■ • ■ + AE P , where the E t are orthogonal idempotents
and each AE t = Xfii + N t where E t Ni — N t . Show that
e A = 2 e^e^Ei.
% 7 I Systems of Linear Differential Equations
Most of this section can be read with background material from Chapter I,
the first four sections of Chapter II, and the first five sections of Chapter III.
However, the examples and exercises require the preceding section, Section 5.
Theorem 7.3 requires Section 1 1 of Chapter V. Some knowledge of the theory
of linear differential equations is also needed.
We consider the system of linear differential equations
dx n
x i = — 1 = JfliXO^. i = 1, 2, . . . , n (7.1)
at t=i
where the a tj (t) are continuous real valued functions for values of t, t x < t <
t 2 . A solution consists of an rctuple of functions X(t) = (x 1 (0, . . . , x n (t))
satisfying (7.1). It is known from the theory of differential equations that
corresponding to any ntuple X = (x 10 , x 20 , . . . , x n0 ) of scalars there is a
unique solution X(t) such that X(t ) — X . The system (7.1) can be written
in the more compact form
X = AX, (7.2)
where ^(0 = (x^t), ... , x n (t)) and A = [a ti (t)].
The equations in (7.1) are of first order, but equations of higher order
can be included in this framework. For example, consider the single «th
order equation
d n x d n ~ x x dx , „
This equation is equivalent to the system
Xy = X 2
JCn —  »£q
X n1 — X n
*n = a n i x n a i x 2  a x 1 . (7.4)
The formula x k = expresses the equivalence between (7.3) and (7.4).
7  Systems of Linear Differential Equations 279
The set of all realvalued differentiable functions denned on the interval
[t lt t 2 ] forms a vector space over R, the field of real numbers. The set V
of ntuples X(t) where the x k {t) are differentiable also forms a vector space
over R. But V is not of dimension n over R; it is infinite dimensional. It is
not profitable for us to consider V as a vector space over the set of differenti
able functions. For one thing, the set of differentiable functions does not
form a field because the reciprocal of a differentiable function is not always
differentiable (where that function vanishes). This is a minor objection because
appropriate adjustments in the theory could be made. However, we want the
^i j • dax
operation of differentiation to be a linear operator. The condition — =
chc
a — requires that a must be a scalar. Thus we consider V to be a vector space
over R.
Statements about the linear independence of a set of «tuples must be
formulated quite carefully. For example.
^(0 = 0,0), *.(0 = (*,0)
are linearly independent over R even though the determinant
1
=
for each value of t. This is because X x {t ) and X 2 (t ) are linearly dependent
for each value of t , but the relation between X^) and X 2 (t ) depends on
the value of t . In particular, a matrix of functions may have n independent
columns and not be invertible and the determinant of the matrix might be 0.
However, when the set of ntuples is a set of solutions of a system of differ
ential equations, the connection between their independence for particular
values of t and for all values of t is closer.
Theorem 7.1. Let t be any real number, t x < t < t 2 , and let {X^t), . . . ,
X m (t)} be a finite set of solutions o/(7.1). A necessary and sufficient condition
for the linear independence of {X^t), . . . , X m (t)} is the linear independence
ofiX^t,), . . . , X m (t )}.
proof. If the set {X^t), . . . , X m (t)} is linearly dependent, then certainly
(XiOo), . . . , X m (t )} is linearly dependent. On the other hand, suppose
there exists a nontrivial relation of the form
m
Zc k X k (t ) = 0. (7.5)
Consider the function X(t) = £™ =1 c k X k (t). X(t) is a solution of (7.1).
But the function Y(t) = is also a solution and satisfies the condition
Y(t Q ) = = X(t ). Since the solution is unique, we must have X(t) = 0.
280 Selected Applications of Linear Algebra  VI
It follows immediately that the system (7.1) can have no more than n
linearly independent solutions. On the other hand, for each k there exists
a solution satisfying the condition X k (t ) = (d lk , . . . , d nk ). It then follows
that the set {X x {t), . . . , XJfj) is linearly independent. □
It is still more convenient for the theory we wish to develop to consider
the differential matrix equation
F= AF (7.6)
where F(t) = [/^(Ol is an n X w matrix of differentiable functions and
F = [fljit)]. If an F is obtained satisfying (7.6), then each column of F
is a solution of (7.2). Conversely, if solutions for (7.2) are known, a solution
for (7.6) can be obtained by using the solutions of (7.2) to form the columns
of F. Since there are n linearly independent solutions of (7.2) there is a
solution of (7.6) in which the columns are linearly independent. Such a
solution to (7.6) is called a. fundamental solution. We shall try to find the
fundamental solutions of (7.6).
The differential and integral calculus of matrices is analogous to the cal
culus of real functions, but there are a few important differences. We have
already defined the derivative in the statement of formula (7.6). It is easily
seen that
l (F + G) = ^ + ^, (7.7)
at at at
and
1 (FG) = — G + F — . (7.8)
dt dt dt
Since the multiplication of matrices is not generally commutative, the
order of the factors in (7.8) is important. For example, if F has an inverse
F 1 , then
= £ (f x F) = ^^ F + F' 1 — . (7.9)
dt dt dt
Hence,
^ F1 > = p^ — F 1 . (7.10)
dt dt
This formula is analogous to the formula (dx^/dt) = —x~\dx\dt) in
elementary calculus, but it cannot be written in this simpler form unless F
and dFjdt commute.
Similar restrictions hold for other differentiation formulas. An important
example is the derivative of a power of a matrix.
dFl^dl F + F dF (7n)
dt dt dt
7  Systems of Linear Differential Equations 281
Again, this formula cannot be further simplified unless Fand dF\dt commute.
However, if they do commute we can also show that
*£l = kF ^i4L m (7.12)
dt dt
Theorem 7.2. If F is a fundamental solution of (7.6), then every solution
G of (7.6) is of the form G = FC where C is a matrix of scalar s.
proof. It is easy to see that if F is a solution of (7.6) and C is an n x n
matrix of scalars, then FC is also a solution. Notice that CF is not necessarily
a solution. There is nothing particularly deep about this observation, but it
does show, again, that due care must be used.
Conversely, let F be a fundamental solution of (7.6) and let G be any
other solution. By Theorem 7.1, for any scalar t F(t ) is a nonsingular
matrix. Let C = Fi^G^). Then H = FC is a solution of (7.6). But
H(t Q ) = F(t )C = G(t ). Since the solution satisfying this condition is
unique we have G = H = FC. O
Let
Jto
B(t) = A(s) ds (7.13)
where B(t) = [& w (f)l and b u {t) — JJ a tj (s) ds. We assume that A and B
dB k
commute so that — — = kAB* 1 . Consider
dt
e B =lf~. (714)
7c=o k !
Since this series converges uniformly in any finite interval in which all the
elements of B are continuous, the series can be differentiated term by term.
Then
df_ =  /cAB^ 1 = ^ & = AeB . (7 15)
dt k=o k\ k=o k !
that is, e B is a solution of (6.6). Since e B{to) = e° = I, e B is a fundamental
solution. The general solution is
F(t) = e mt) C. (7.16)
In this case, under the assumption that A and B commute, e B is also a
solution of F = FA.
As an example of the application of these ideas consider the matrix
1 It
A =
It 1
282
and take t = 0. Then
Selected Applications of Linear Algebra  VI
B =
■f t 2 '
t 2 t
It is easily verified that A and B commute. The characteristic polynomial
of B is (x  t  t 2 )(x  t + t 2 ). Taking X x = t + t 2 and X 2 = t  t 2 , we
have e x (x) = — (x — t + t 2 ) and e 2 (x) = — — (x — t — t 2 ). Thus,
1
I = 2
"i r
i
+ 
i i_
2
1
1
is the corresponding resolution of the identity. By (6.28) we have
e t+t*
2
"1 f
.1 1.
tt 2
1 
.1
r
1
V+' 2 + e** 2 e t+t *  e** 2 '
2
J
^,
tt 2 e t+i
2 +^ 2 J
Since e B is a fundamental solution, the two columns of e B are linearly
independent solutions of (7.2), for this A.
If A is a matrix of scalars, B = (t — t )A certainly commutes with A
and the above discussion applies. However, in this case the solution can be
obtained more directly without determining e B . If C = (c l5 . . . , c„),
Ci e R, represents an eigenvector of A corresponding to I, that is, AC = AC,
then BC = (t — t )AC so that {t — t Q )X is an eigenvalue of B and C represents
an eigenvector of B. Then
X(t) = (c^™^, c 2 e Mt  to \
> £«£
Mtt )
(7.17)
is a solution of (7.2), as can easily be verified. Notice that X(t ) = C.
Conversely, suppose that X{t) is a solution of (7.2) such that X(t ) = C
is an eigenvector of A. Since X(t) is also a solution of (7.2),
Y{t)=XXX (7.18)
is also a solution of (7.2). But Y = X  XX = AX  XX = (A  XI)X for
all t, and F(f ) = (A — XI)C = 0. Since the solution satisfying this condition
is unique, Y(t) = 0, that is, X = XX. This means each x,(f) must satisfy
x t {t) = Xx,{t), Xj (t ) = c,. (7.19)
Thus X(t) is given by (7.17). This method has the advantage of providing
solutions when it is difficult or not necessary to find all eigenvalues of A.
7  Systems of Linear Differential Equations 283
Return now to the case where A is not necessarily a matrix of constants.
Let G be a fundamental solution of
G = A T G. (7.20)
This system is called the adjoint system to (7.6). It follows that
WL^D. = q t F + G T F = {A T G) T F + G T AF (7.21)
dt
= G T (A)F + G T AF = 0.
Hence, G^F = C, a matrix of scalars. Since F and G are fundamental
solutions, C is nonsingular.
An important special case occurs when
A T = A. (7.22)
In this case (7.6) and (7.20) are the same equation and the equation is said
to be selfadjoint. Then F is also a solution of the adjoint system and we see
that F T F = C. In this case C is positive definite and we can find a non
singular matrix D such that D T CD = /. Then
D T F T FD = I = (FD) T (FD). (7.23)
Let FD = N. Here we have N T N = / so that the columns of N are ortho
normal solutions of (7.2).
Conversely, let the columns of F be an orthonormal set of solutions of
(7.2). Then
= d(F^F) = p Tp + pTf = {AF) T F + f t af = F T( A T + A)F (7 24)
dt
Since F and F T have inverses, this means A T + A = 0, so that (7.6) is
selfadjoint. Thus,
Theorem 7.3. The system (7.2) has an orthonormal set of n solutions if
and only if A T = —A. □
BIBLIOGRAPHICAL NOTES
E. A. Coddington and N. Levinson, Theory of Ordinary Differential Equations; F. R.
Gantmacher, Applications of the Theory of Matrices; W. Hurewicz, Lectures on Ordinary
Differential Equations; and S. Lefschetz, Differential Equations, Geometric Theory, contain
extensive material on the use of matrices in the theory of differential equations. They differ
in their emphasis, and all should be consulted.
284 Selected Applications of Linear Algebra  VI
EXERCISES
1. Consider the matric differential equation
"1 2
X =
2 1
X = AX.
Show that (1,1) and (1, 1) are eigenvectors of A. Show that X x = (—<?<**<>>,
^(t*o>) and X 2 = (e 3(t_ 'o> 5 c 3(t— « )) are solutions of the differential equation.
Show that
" _ e Ut ) g3(t*o>~
F =
e «< ) e 3(tt )
is a fundamental solution of F = AF.
Go through a similar analysis for the equation X = AX for each matrix A given
in Exercises 1 and 2 of Section 6.
2. Let /I = IE + N where iV is idempotent and EN = NE = N. Show that
e At = I  E + e u e Nt E.
3. Let Y = e Nt where TV is nilpotent. Show that
Y = Ne Nt .
(This is actually true for any scalar matrix N. However, the fact that N is nilpotent
avoids dealing with the question of convergence of the series representing e N and
the question of differentiating a uniformly convergent series term by term.)
4. Let A = AE X + ■ ■ ■ + AE P where the E { are orthogonal idempotents and
each AE t is of the form AE t = \E t + N f where ^iV* = N& = N t and N t is
nilpotent. Show that v
B =e At =£ e Xit e N ^Ei.
Show that i=1
B = AB.
8 I Small Oscillations of Mechanical Systems
This section requires a knowledge of real quadratic forms as developed
in Section 10 of Chapter IV, and orthogonal diagonalization of symmetric
matrices as achieved in Section 1 1 of Chapter V. If the reader is willing to
accept the assertions given without proof, the background in mechanics
required is minimal.
Consider a mechanical system consisting of a finite number of particles,
{/*!, . . . , P r }. We have in mind describing the motions of such a system
for small displacements from an equilibrium position. The position of each
particle is specified by its three coordinates. Let (x 3i _ 2 , x 3i _ t , x 3i ) be the
coordinates of the particle iV Then (x lt x tt . . . , x 3r ) can be considered to
8  Small Oscillations of Mechanical Systems 285
be the coordinates of the entire system since these 3r coordinates specify
the location of each particle. Thus we can represent the configuration of a
system of particles by a point in a space of dimension n = 3r. A space of
this kind is called a phase space.
The phase space has not been given enough structure for us to consider
it to be a vector space with any profit. It is not the coordinates but the
small displacements we have to concentrate our attention upon. This is a
typical situation in the applications of calculus to applied problems. For
example, even though y =f(x lt ...,*„) may be a transcendental function
of its variables, the relation between the differentials is linear,
d 9 &d* l + p!dx t + +2!dx.. (8.1)
dx x ox 2 ox n
In order to avoid a cumbersome change of notation we consider the co
ordinates (x lt . . . , x n ) to be the coordinates of a displacement from an
equilibrium position rather than the coordinates of a position. Thus the
equilibrium point is represented by the origin of the coordinate system.
We identify these displacements with the points (or vectors) in an «dimen
sional real coordinate space.
The potential energy V of the system is a function of the displacements,
V = V(x x , . . . , x n ). The Taylor series expansion of V is of the form
IdV dV
V = V + — x x + •••+— x r
ox„
1/dV , . d 2 v 2 , . d 2 v . 2 , „ d 2 v
2 X n
i
d 2 V
+  I X X x + : X 2 + +   X n h L X X X 2
2\dx 2 dx 2 dx 2 dx x dx 2
+ . . . + 2 — x n _ x x n \ + • ■ ■
(8.2)
We can choose the level of potential energy so that it is zero in the equi
librium position; that is, V = 0. The condition that the origin be an
dV dV
equilibrium position means that — = • • • =  — = 0. If we let
dx x dx,
d 2 V d 2 V
= a i4 =
dx, dx, dx d dx.
= a M , (8.3)
then
V = \ 2 WW +••■■ (8.4)
286 Selected Applications of Linear Algebra  VI
If the displacements are small, the terms of degree three or more are small
compared with the quadratic terms. Thus,
V = ; I *i*n*i (85)
is a good approximation of the potential energy.
We limit our discussion to conservative systems for which the equilibrium
is stable or indifferent. If the equilibrium is stable, any displacement must
result in an increase in the potential energy; that is, the quadratic form in
(8.5) must be positive definite. If the equilibrium is indifferent, a small
displacement will not decrease the potential energy; that is, the quadratic
form must be nonnegative semidefinite.
The kinetic energy Tis also a quadratic form in the velocities,
T = \ I iib u x t . (8.6)
In this case the quadratic form is positive definite since the kinetic energy
cannot be zero unless all velocities are zero.
In matrix form we have
V = \X T AX, (8.7)
and
T = hX T BX, (8.8)
where A = [a it ], B = [b it \ t X = (x x , . . . , x n ), and X = (x lt ... , x n ).
Since B is positive definite, there is a nonsingular real matrix Q such
that Q T BQ = I. Since Q T AQ = A' is symmetric, there is an orthogonal
matrix Q' such that Q' T A'Q' = A" is a diagonal matrix. Let P = QQ'.
Then
P T AP = Q' T Q T AQQ' = Q' T A'Q' = A" (8.9)
and
P T BP = Q' T Q T BQQ' = Q' T IQ' = Q' T Q' = I. (8.10)
Thus P diagonalizes A and B simultaneously. (This is an answer to Exercise 3 ,
Chapter V 11.)
If we set Y = P' 1 X, then (8.7) and (8.8) become
and
V = lY T A"Y = \j l a i y i t , (8.11)
2i=l
T = \Y T Y=\j,y?, (8.12)
2i=l
where a t is the element in the ith place of the diagonal of A".
8  Small Oscillations of Mechanical Systems 287
In mechanics it is shown that the Lagrangian L= T — V satisfies the
differential equation
d_/dL\ _dL = 0i j = 1, . . . , n . (8.13)
dt\dyj dyi
For a reference, see Goldstein, Classical Mechanics, p. 18. Applied to
(8.11) and (8.12), this becomes
ft + ««y< = 0, i = 1,...,/i. (8.14)
If A is positive definite, we have a t > 0. If A is nonnegative semidefinite,
we have a t > 0. For those a t > 0, let a f = a>< 2 where eo* > 0. The solutions
of (8.14) are then of the form
y .(t) = Cj cos {(o f t + e t ), j=l, ... ,n. (8.15)
p = [p if ] is the matrix of transition from the original coordinate system
with basis A = {a l5 . . . , aj to a new coordinate system with basis B =
{fi lt ..., p n }; that is, & = 27=i/>» a r Thus '
*i(0 = 2p< t 9#) = IPijCf cos ((Oft + 0,) (8.16)
0=1 3=1
in the original coordinate system. If the system is displaced to an initial
position in which
y k (0) =l,t/,(0) = for j * k, (8.17)
&(0) = 0, j=l,...,n,
then
and
or
y k (t) = cos (o k t, yf(t) = for j 5* k, (8.18)
x A t ) = Pik cos °V> ( 8  19 )
*(0 = (Pi^P2 k , • • • > />»*) cos w *' ( 8  2 °)
8 fc is represented by (p lk ,p 2k , ... , p nk ) in the original coordinate system.
This wtuple represents a configuration from which the system will vibrate
in simple harmonic motion with angular frequency (o k if released from rest.
The vectors {&, . . . , £„} are called the principal axes of the system. They
represent abstract "directions" in which the system will vibrate in simple
harmonic motion. In general, the motion in other directions will not be
simple harmonic, or even harmonic since it is not necessary for the w, to be
commensurable. The coordinates (y lf . . . , y n ) in this coordinate system
are called the normal coordinates.
We have described how the simultaneous diagonalization of A and B
288 Selected Applications of Linear Algebra  VI
can be carried out in two steps. It is often more convenient to achieve the
diagonalization in one step. Consider the matric equation
(A  XB)X = 0. (8.21)
This is an eigenvalue problem in which we are asked to find a scalar X for
which the equation has a nonzero solution. This means we must find a
A for which
det(A  XB) = 0. (8.22)
Using the matrix of transition P given above we have
= det P T ■ det(A  XB) • det P = det(P T (A  XB)P)
= det(P T AP  XP T BP) = det{A"  XI). (8.23)
Since A" is positive definite or nonnegative semidefinite the eigenvalues
are >0. In fact, these eigenvalues are the a t of formula (8.11).
Let X x and X 2 be eigenvalues of equation (8.21) and let X x and X 2 be
corresponding solutions. If X x ^ X 2 , then
Aj AX 2 = Aj {X 2 BX 2 ) = X 2 X X BX 2
= (A T X 1 ) T X 2 = (AX 1 ) T X 2
= (X l BX 1 ) T X 2 = X X X X T BX 2 . (8.24)
Thus,
X t T BX 2 = 0. (8.25)
This situation is described by saying that X x and X 2 are orthogonal with
respect to B. This is the same meaning given to this term in Chapter Vl
where an arbitrary positive definite quadratic form was selected to determine
the inner product.
This argument shows that if the eigenvalues of (8.21) are distinct, then
there exists an orthonormal basis with respect to the inner product defined
by B. We must show that such a basis exists even if there are repeated
eigenvalues. Let a be the linear transformation represented by A" with
respect to the basis 8, that is, cr(^) = a^. Let X t be the representation of
fa with respect to the basis A, Xj = (p Xj , . . . ,p ni ). The matrix representing
a with respect to A is PA"P~ X . Thus,
PA"P~ 1 X j = djX,. (8.26)
Then
AX, = (P T ) 1 P T APP~ 1 X j
= (P T ) 1 A"P~ 1 X j
= {P T ) 1 P X PA"P X X J
= (P^P^X,
= a t BX,. (8.27)
8 I Small Oscillations of Mechanical Systems 289
Fig. 5
mi
Since (A — a^Xj = 0, we see that a t is an eigenvalue of (8.21) and X t
is a corresponding eigenvector. Since the columns of P are the X iy the
condition (8.10) is equivalent to the statement that the X, are orthonormal
with respect to the inner product defined by B.
We now have two related methods for finding the principal axes of the
given mechanical system: diagonalize B and A in two steps, or solve
the eigenvalue problem (8.21). Sometimes it is not necessary to find all the
principal axes, in which case the second method is to be preferred. Both
methods involve solving an eigenvalue problem. This can be a very difficult
task if the system involves more than a few particles. If the system is highly
symmetric, there are other methods for finding the principal axes. These
methods are discussed in the next sections.
We shall illustrate this discussion with a simple example. Consider the
double pendulum in Fig. 5. Although there are two particles in the system,
the fact that the pendulum rods are rigid and the system is confined to a
plane means that the phase space is only 2dimensional. This example also
illustrates a more general situation in which the phase space coordinates are
not rectangular.
The potential energy of the system is
V = gnhil  I cos xj + gm 2 (2l  I cos x x — I cos x 2 )
= gl[(mi + 2w 2 ) — K + m 2 ) cos x x — m 2 cos x 2 ].
The quadratic term is
v = \ K m i + m 2>i 2 + rn 2 x 2 2 }.
290
Selected Applications of Linear Algebra  VI
r*i
2x1
Fig. 6
The kinetic energy is
T = ^ntxilxj) 2 + \m 2 l 2 [x x 2 + x 2 2 + 2 cos (x x + x 2 )x x x 2 ].
The quadratic term is
T = £/ 8 [(/»i + Wa)^ 2 + 2m 2 x x x 2 + m 2 i 2 2 ].
To simplify the following computation we take m 1 = 3 and m 2 = 1 . Then
we must simultaneously diagonalize
A=gl
Solving the equation (8.22), we find X x = (2g/3l), A 2 = 2g/l. This gives
o>i = y/2g/3l, co 2 = \2g/l. The coordinates of the normalized eigenvectors
are X x = — = (1, 2), X 2 = — (1, —2). The geometrical configuration for
the principal axes are shown in Fig. 6.
The idea behind the concept of principal axes is that if the system is started
from rest in one of these two positions, it will oscillate with the angles x x
and x 2 remaining proportional. Both particles will pass through the vertical
line through the pivot point at the same time. The frequencies of these two
4 0"
B = I 2
4 1
1
i
1 1
8  Small Oscillations of Mechanical Systems 291
modes of oscillation are incommensurable, their ratio being %/3. If the
system is started from some other initial configuration, both modes of
oscillation will be superimposed and no position will ever be repeated.
BIBLIOGRAPHICAL NOTES
A simple and elegant treatment of the physical principles underlying the theory of
small oscillations is given in T. von Karman and M. A. Biot, Mathematical Methods in
Engineering.
EXERCISES
Consider a conservative mechanical system in the form of an equilateral triangle
with mass M at each vertex. Assume the sides of the triangle impose elastic forces
according to Hooke's law ; that is, if j is the amount that a side is stretched from an
equilibrium length, then ks is the restoring force, and ks 2 /2 = ft ku du is the
potential energy stored in the system by stretching the side through the distance s.
Assume the constant k is the same for all three sides of the triangle. Let the triangle
be placed on the x, y coordinate plane so that one side is parallel to the a>axis.
Introduce a local coordinate system at each vertex so that the that the coordinates
of the displacements are as in Fig. 7. We assume that displacements perpendicular
to the plane of the triangle change the lengths of the sides by negligible amounts
so that it is not necessary to introduce a zaxis. All the following exercises refer
to this system.
1. Compute the potential energy as a function of the displacements. Write
down the matrix representing the quadratic form of the potential energy in the
given coordinate system.
2. Write down the matrix representing the quadratic form of the kinetic energy
of the system in the given coordinate system.
yz
*3
A / \ k
t
yi
— >
X\ X2
Fig. 7
292 Selected Applications of Linear Algebra  VI
3. Using the coordinate system,
z x =x 2  x z y' 1 =y 2  y z
*2= X 3 X l 2/2=2/3— Vl
x z = x * y' z = y%
The quadratic form for the potential energy is somewhat simpler. Determine the
matrix representing the quadratic form for the potential energy in this coordinate
system.
4. Let the coordinates for the displacements in the original coordinate system
be written in the order, (x lt y x , x 2 , y 2 , x 3 , y 3 ). Show that (1, 0, 1, 0, 1, 0) and
(0, 1, 0, 1, 0, 1) are eigenvectors of the matrix V representing the potential energy.
Give a physical interpretation of this observation.
9 I Representations of Finite Groups by Matrices
For background the material in Chapters I, II, III, and V is required,
except for Section 8 of Chapter III. A knowledge of elementary group
theory is also needed. Appreciation of some of the results will be enhanced
by a familiarity with Fourier transforms.
The theory of representations of finite groups is an elegant theory with its
own intrinsic interest. It is also a finite dimensional model of a number
of theories which outwardly appear to be quite different; for example,
Fourier series, Fourier transforms, topological groups, and abstract harmonic
analysis. We introduce the subject here mainly because of its utility in finding
the principal axes of symmetric mechanical systems.
We have to assume a modest knowledge of group theory. In order to be
specific about what is required we state the required definitions and theorems.
We do not give proofs for the theorems. They are not difficult and can be
considered to be exercises, or their proofs can be found in the first few pages
of any standard treatment of group theory.
Definition. A group G is a set of elements in which a law of combination is
defined having the following properties :
(1) If a, b e G, then ab is uniquely defined by the law of combination and
is an element of G.
(2) (ab)c = a(bc), for all a, b, c e G.
(3) There exists an element e e G, called the unit element, such that
ea = ac = a for each a e G.
(4) For each a e G there exists an element a 1 , called the inverse of a,
such that a _1 a = aa _1 = e.
Although the law of combination is written multiplicatively, this does
not mean that it has anything to do with multiplication. For example, the
9  Representations of Finite Groups by Matrices 293
field of rational numbers is a group under addition, and the set of positive
rational numbers is a group under multiplication. A vector space is a group
under vector addition. If the condition
(5) ab = ba
is also satisfied, the group is said to be commutative or abelian.
The number of elements in a group is called the order of the group. We
restrict our attention to groups of finite order.
A subset of a group which satisfies the group axioms with the same law of
combination is called a subgroup. Let G be a group and S a subgroup of G.
For each a £ G, the set of all b £ G such that a _1 b £ S is called a left coset of
S defined by a. By aS we mean the set of all products of the form ac where
c e S. Then a _1 b £ S is equivalent to the condition b £ aS; that is, aS is the
left coset of S determined by a. Two left cosets are equal, aS = bS, if and
only if S = a^bS or a _1 b £ S; that is, if and only if b £ aS. The number of
elements in each coset of S is equal to the order of S. A right coset of S is of
the form Sa.
Theorem 9.1. If G is of finite order and S is a subgroup of G, the order
of S divides the order of G. □
If S is a subgroup such that its right and left cosets are equal — that is,
aS = Sa for each a £ G — then S is said to be an invariant or normal subgroup
of G. If S is normal, then (aS)(bS) = a(Sb)S = a(bS)S = (ab)SS = abS so
that the product of two cosets is a coset.
Theorem 9.2. If S is a normal subgroup ofG, the cosets of S form a group
under the law of combination (aS)(bS) = abS. □
If S is a normal subgroup of G, the group of cosets is called the factor
group of G by S and is denoted by G/S.
If G x and G 2 are groups, a mapping /of G x into G 2 is called a homomorphism
if /(ab) =/(a)/(b) for all a, b £ G x . Notice that the law of combination
on the left is that in G l5 while the law of combination on the right is that in G 2 .
If the homomorphism is onetoone, it is called an isomorphism. If e is the
unit element in G 2 , the set of all a £ G x such that/(a) = e is called the kernel
of the homomorphism.
Theorem 9.3. (The homomorphism theorem). If G 2 is the image of G 2
under the homomorphism f and K is the kernel off, then G 2 is isomorphic
to GJK, where f (a) £ G 2 corresponds to aK £ GJK. □
An isomorphism of a group onto itself is called an automorphism. For
each fixed a £ G, the mapping / a of G onto itself defined by / a (x) = a _1 xa
is an automorphism. It is called an inner automorphism.
294 Selected Applications of Linear Algebra  VI
Two group elements b x and b 2 are said to be conjugate if there is an a e G
such that a _1 b 1 a = b 2 . For each b e G, the set of conjugates of b is called
the conjugate class determined by b. The conjugate class determined by b
is the set of all images of b under all possible inner automorphisms. The
normalizer N b of b is the set of all a such that the inner automorphism / a
leaves b fixed.
Theorem 9.4. The normalizer N b is a subgroup of G. The number of con
jugates ofb is equal to the number of left cosets ofN b . The number of elements
in conjugate class is a divisor of the finite order of the group. □
We have already met several important groups. The set of all n x n
matrices with nonzero determinant forms a group under matrix multipli
cation. This group is infinite. There are several important subgroups;
the matrices with determinant ±1, the orthogonal matrices, etc. These
are also infinite. There are also a large number of finite subgroups. Given
a finite group, representation theory seeks to find groups of matrices which
are models, in some sense, of the given group.
Definition. Let G be a given finite group. A representation of G is a group
D(G) of square matrices (under the operation of matrix multiplication)
with a homomorphism mapping G onto D(G). Notice that the homo
morphism must be specified and is part of the representation. If the mapping
is an isomorphism, the representation is said to be faithful. The matrices
in D(G) represent linear transformations on a vector space V over C. The
corresponding linear transformation also form a group homomorphic to G,
and we also say that this group of transformations represents G. The
dimension of V is called the dimension of the representation.
If a e G, we denote by D(a) the matrix in D(G) corresponding to a under
the homomorphism of G onto D(G). <x a will denote the corresponding linear
linear transformation on V.
Since <r a = a & a b ia h and the rank of a product is less than or equal to
the rank of any factor, we have piaj < p(o h ). Similarly, we must have
p(°b) ^ P( a t)> an d hence the ranks of all matrices in D(G) must be equal.
Let their common rank be r. Then a a (V) = S a is of dimension r, and
a &2 (V) = o a (a & (V)) = a & (S & ) is also of dimension r. Since o^SJ c: S a ,
we have cr a (S a ) = S a . Also S a = <r a (V) = a^a^a^V)) <= <r b (V) = S b .
Similarly, S b <= S a so that S a = S b . This means that all linear transformations
representing G are automorphisms of some subspaces S of dimension r in
V, and cr a (V) = S for all a e G. Thus, we may as well restrict our attention
to S, and in the future we assume, without loss of generality, that the matrices
and transformations representing a group are nonsingular, that is, S = V.
A subspace U <= V such that a a (U) <= U for all a e G is called an invariant
9  Representations of Finite Groups by Matrices 295
subspace of V under D(G). Since <r a is nonsingular, actually <r & (U) = U.
This means that U is also a representation space of G over C. If V has a
proper invariant subspace we say that V and the representation are reducible.
Otherwise the representation and the space are said to be irreducible. If
there is another proper invariant subspace W such that V = U ® W, we say
that V and the representation are completely reducible. Notice that if U x
and U 2 are invariant under D(G), then U 1 n U 2 is also invariant under D{G).
Theorem 9.5. For a finite group G a reducible representation is completely
reducible.
proof. Let U be an invariant subspace. Let T be any subspace com
plementary to U, and let n be the projection of V onto U along T. Since
<7 a (Lf) = U, itojt = (r a 7r. Thus, in a complicated product of tr's and 7r's,
all 7r's to the left of the last one can be omitted; for example, TTa t jra h 7ra c =
cr a (r a 7r<r c . Consider
T =  2 ffaVTTffa, (9.1)
g a
where g is the order of G, and the sum is taken over all a e G.
An expression like (9.1) is called a symmetrization of it. t has the important
property that it commutes with each a a whereas n might not. Specifically,
1 v lv
TOr b =  2, OaiTrOgCTb =  2, ffb^b^^Cab
g • g a
g a
The reasoning behind this conclusion is worth examining in detail. If
G = {a l5 a 2 , . . . , a ff } is the list of the elements in G written out explicitly
and b is any element in G, then {axb, a 2 b, . . . , a ff b} must also be a list (in a
different order) of the elements in G because of axioms (1) and (4). Thus, as a
runs through the group, ab also runs through the group and ^ a a (a.b) l7Ta &b =
2a a & l7T(J & Al so notice that this conclusion does not depend on the con
dition that 77 is a projection. The symmetrization of any linear transformation
would commute with each <r a .
Now,
TTT =  ^ OalTTffaTT = ~ 2 <J 2T ia ^' IT = ~ 2 "" = ""» ( 9  3 )
g * g a g a
and
77T = — 2 TTOaiTTCa = — 2 a X~ 17T<J * = T  (94)
g a g a
Among other things, this shows that t and 7r have the same rank. Since
T (V) = 7tt(V) c U, this means that t(V) = U. Then t 2 = t(ttt) = (ttt)t =
7tt = t, so that t is a projection of V onto U.
D(a) =
(9.5)
296 Selected Applications of Linear Algebra  VI
Let W be the kernel of t. Then, for each a g G, ra & (W) = a a r(W) =
°a(0) — 0. Thus o a (W) c= W and W is an invariant subspace. Finally,
it is easily seen that V = U © W. Thus, the representation is completely
reducible. □
The importance of complete reducibility is that the representation on V
induces two representations, one on U and one on W. If a basis {a l5 . . . ,
a r , /?!, . . . , fi n _ r } of V is chosen so that {a x , . . . , a r } is a basis of U and
{&, . . . , /? n _ r } is a basis of W, then Z)(a) is of the form
"^(a) '
2) 2 (a)
where Z> x (a) is an r x r matrix and Z> 2 (a) is an n — r x n — r matrix.
Z) 1 (a) represents a^rr on (/, and Z> 2 (a) represents <r a (l — 77) on W. The
set X) 1 (G) = {D x (2l)  a g G} is a representation of G on L/, and D 2 (G) =
(Z) 2 (a) I a e G} is a representation of G on W. We say that D(G) is the
direct sum of Z^G) and D 2 (G) and we write D(G) = D X (G) + Z) 2 (G).
If either the representation on U or the representation on W is reducible,
we can decompose it into a direct sum of two others. We can proceed
in this fashion step by step, and the process must ultimately terminate
since at each step we obtain subspaces of smaller dimensions. When that
point is reached we will have decomposed D(G) into a direct sum of irreduc
ible representations. If U is an invariant subspace obtained in one decom
position of V and U' is an invariant subspace obtained in another, then
U n U' is also invariant. If U and W are both irreducible, then either
U n U' = {0} or U = U'. Thus, the irreducible subspaces obtained are
unique and independent of the particular order of steps in decomposing
V. We see, then, that the irreducible representations will be the ultimate
building blocks of all representations of finite groups.
Although the irreducible invariant subspaces of V are unique, the matrices
corresponding to the group elements are not unique. They depend on the
choices of the bases in these subspaces. We say that two groups of matrices,
D(G) = (Z)(a) I a £ G} and D'{G) = {Z)'(a)  a e G}. are equivalent repre
sentations of G if there is a nonsingular matrix P such that D'(a) = P _1 Z)(a)P
for every a g G. In particular, if the two groups of matrices represent the
same group of linear transformations on V, then they are equivalent and P
is the matrix of transition. But the definition of equivalence allows another
interpretation.
Let V and V be vector spaces over C both of dimension n. Let/ be a one
toone linear transformation mapping V onto V, and let P be the matrix
representing/. We can define a linear transformation r a on V by the rule
r & =f^f. (9.6)
9  Representations of Finite Groups by Matrices 297
It is easy to show that the set {r a } defined in this way is a group and that it
is isomorphic to {a & }. The groups of matrices representing these linear
transformations are also equivalent.
If a representation D(G) is given arbitrarily, it will not necessarily look
like the representation in formula (9.5). However, if D(G) is equivalent
to a representation that looks like (9.5), we also call that representation reduc
ible. The importance of this notion of equivalence is that there are only
finitely many inequivalent irreducible representations for each finite group.
Our task is to prove this fact, to describe at least one representation for each
class of equivalent irreducible representations, and to find an effective
procedure for decomposing any representation into its irreducible com
ponents.
Theorem 9.6 {Schur's lemma). Let D\G) and D 2 (G) be two irreducible
representations. If T is any matrix such that TD 1 ^) = D 2 (n)Tfor all a e G,
then either T = or T is nonsingular and D\G) and D 2 (G) are equivalent.
proof. Let V x of dimension m be the representation space of D\G),
and let V 2 of dimension n be the representation space of D 2 (G). Then T
must have n rows and m columns, and so may be thought of as representing
a linear transformation / of V x into V 2 . Let a x a be the linear transformation
on V x represented by D^a) for a e G, and let <r 2 a be the linear transformation
on V 2 represented by D 2 (a). Since fa la = o 2 Jf, ^J/TO] =/[<r 1> ,(V 1 )] =
/(V x ) so that f(V x ) is an invariant subspace of V 2 . Since V 2 is irreducible,
either/(Vi) = and hence T = 0, or else/(V 1 ) = V 2 .
If a efi(0), then/[or la ( a )] = a 2 Jf(a.)] = (T 2 , a (0) = 0, so that (r 1>B (a) e
/HO). Thus /"HO) is an invariant subspace of V v Since V x is irreducible,
either /HO) = 0, or else/ x (0) = V x in which case T = 0.
Thus, either /(V x ) = 0, /"HO) = V lt and T = 0; or else f(Vd = V 2 ,
/i(0) = 0, and T is nonsingular. In the latter case the representations are
equivalent. □
Theorem 9.7. Let D{G) be an irreducible representation over the complex
numbers. If T is any matrix such that TD(n) = D(a)T for all a 6 G, then
T = XI where X is a complex number.
proof. Let X be an eigenvalue of T. Then (T  XI)D{a) = D(n)(T XI)
and, by Theorem 9.6, T — XI is either or nonsingular. Since T — XI
is singular, T = XL o
Theorem 9.8. If D(G) is an irreducible representation such that any two
matrices in D(G) commute, then all the matrices in D(G) are of first order.
proof. By Theorem 9.7, all matrices in D(G) must be of the form XL
But a set of matrices of this form can be irreducible only if all the matrices
are of first order. □
298 Selected Applications of Linear Algebra  VI
With the proof of Schur's lemma and the theorems that follow im
mediately from it, the representation spaces have served their purpose.
We have no further need for the linear transformations and we have to be
specific about the elements in the representing matrices. Let Z>(a) = [tf„(a)J.
Theorem 9.9. Every representation of a finite group over the field of
complex numbers is equivalent to a representation in which all matrices are
unitary.
proof. Let D{G) be the representation of G. Consider
H = 2 D(*)*D(n), (9.7)
a
where Z>(a)* is the conjugate transpose of Z>(a). Each Z>(a)*D(a) is a
positive definite Hermitian form and H, as the sum of positive definite
Hermitian forms, is a positive definite Hermitian form. Thus, there is a
nonsingular matrix P such that P*HP = I. But then
(p 1 D(a)P)*(p 1 D(a)P) = P*D(*)*p* 1 p 1 D(*)P
= P*D(n)*H D(a)P
= Wj D(a)*D(b)*D(b)D(a)\p
= P*(2 D(ba)*D(ba)]p
= P*HP = I.
Thus, each P _1 Z)(a)P is unitary, as we wished to show. □
For any matrix A = [a^], S(A) = 2f =x «« is called the trace of A. Since
S(AB) = 2? =1 (Z? =1 a ij b h ) = ^ l (% =1 b ji a ij ) = S(BA), the trace of a
product does not depend on the order of the factors. Thus S{P~ X AP) =
SiAPP' 1 ) = S(A) and we see that the trace is left invariant under similarity.
If D l {G) and D 2 (G) are equivalent representations, then SiD 1 ^)) = S(D 2 (a))
for all a e G. Thus, the trace, as a function of the group element, is the same
for any of a class of equivalent representations. For simplicity we write
S(D(a)) = S D (a) or, if the representation has an index, 5 , (D r (a)) = S r (a).
If a and a' are conjugate elements, a' = b _1 ab, then S^Ca') = S(Z)(b 1 )D(a)
Z)(b)) = ^(a) so that all elements in the same conjugate class of G have the
same trace. For a given representation, the trace is a function of the conjugate
class. If the representation is irreducible, the trace is called a character
of G and is written 5 r (a) = # r (a)
Let D r (G) and Z) S (G) be two irreducible representations. Then consider
the matrix
T = 2 D'itr^XDX*) (9.8)
9  Representations of Finite Groups by Matrices 299
where X is any matrix for which the products in question are defined. Then
TD s (b) = 2 DVP'W^)
a
= D r (b) 2 D'Qr^^XDXab)
a
= D r (b)T. (9.9)
Thus, either D r (G) and D 8 (G) are inequivalent and T = for all X, or
D r {G) and Z^CG) are equivalent. We adopt the convention that irreducible
representations with different indices are inequivalent. If r = s we have
T = col, where co will depend on X. Let Z) r (a) = [o^(a)], T = [f w ], and
let X be zero everywhere but for a 1 in they'th row, fcth column. Then
tu = K/^KOO = for r^s. (9.10)
a
If r = s, then co is a function of r, y, and k and we have
'« = 1 «W» _1 K»(«) = ">* *«• ( 9  n )
a
Notice that co d jk is independent of / and /. But we also have
hi = I «L(a~>L(a)
a
= 2, tf£i(a)<4(a _1 )
a
a
where of u is independent of j and A:. Thus, we see that o) r ik d r u = co r H d ilc =
unless k = j and / = i, in which case to r j:j — of u . But af^ is independent
of / and co r i:j is independent of/ Thus, we may write
Z* r ij (* 1 )a r kl (a) = co r d il d jk . (9.13)
a
In order to evaluate co r set k = j, I = i, and sum overy;
n r
II«r/a 1 M i (a) = "X (914)
where n r is the dimension of the representation D r (G). But since
2"=i fl ii(a 1 ) a ii( a ) is a diagonal element of the product D r (a _1 )Z) r (a) =
Z> r (e) = 7 r we have
n r co r = 2 iX/a 1 )^) = 1 1 = g. (9.15)
a i=l a
300 Selected Applications of Linear Algebra  VI
Thus,
of = £ . (9.16)
n r
All the information so far obtained can be expressed in the single formula
2 arX«">«00 =  <>* <>« <$«• (917)
a n r
Multiply (9.17) by a s lt (b) and sum over /. Then
2 1 aU^aMaM = 2 ~ «A00 *« »« *r»
i =i a 1=1 n r
or (9.18)
2 fl ;xa»b) = ■£ fl^(b) a„ a„.
a n r
In (9.18) set / = /, t = k, and sum over/ and k:
n T n s n T n s „
121 flW»">J*(ab) =22 <40>) ** *«,
3=1 fc=i a i=l fc=l n r
or (9.19)
2/(a"V(ab) = ^ Z s (b)<5 rs .
a n r
In particular, it we take b = e, we have
2 /(a  V(a) =  n s d rs = g d rs . (9.20)
a n r
Actually, formula (9.20) could have been obtained directly from formula
(9.17) by setting i = j, I = k, and summing over/ and k.
Let D(G) be a direct sum of a finite number of irreducible representations,
the representation D r (G) occurring c r times, where c r is a nonnegative
integer. Then
^(a) = 2^ r (a),
r
from which we obtain
2/(a 1 )S 2) (a) = gcv. (9.21)
a
Furthermore,
2S 2) (a 1 )S i) (a)=g2<V 2 , (9.22)
a r
so that a representation is irreducible if and only if
2 SJtT^SJ*) = g. (9.23)
9  Representations of Finite Groups by Matrices 301
In case the representation is unitary, ^(ar 1 ) = a r H (a), so that the relations
(9.17) through (9.23) take on the following forms:
a W,
2 fl»fl«(ab) = ^ a s it (b) d jk d rs , (9.18)'
a n r
Zx r (a)x s (*W = x s Q>)Srs, ( 9  19 )'
a n r
2f(*W(*) = 8*T» ( 9  2 °)'
2/(a)S 1) (a) = gc r , (9.21)'
25^)5^) = g2c r », (922)'
2 S d (h)S d (a) = g if and only if D(G) is irreducible. (9.23)
a
Formulas (9.19)' through (9.23)' hold for any form of the representations,
since each is equivalent to a unitary representation and the trace is the same
for equivalent representations.
We have not yet shown that a single representation exists. We now do
this, and even more. We construct a representation whose irreducible
components are equivalent to all possible irreducible representations. Let
G = {a l9 a 2 , . . . , aj. Consider the set V of formal sums of the form
a = X^ + #2*2 + * • • + X g*g> X i G C  ( 9 24)
Addition is defined by adding corresponding coefficients, and scalar multi
plication is defined by multiplying each coefficient by the scalar factor.
With these definitions, V forms a vector space over C. For each a^eGwe can
define a linear transformation on V by the rule
a 4 (a) = SifoHi) + x 2 (*&) + • • • + *„(*&„)> (925)
^ induces a linear transformation that amounts to a permutation of the
basis elements. Since (a i a i )(a) = a^a^a)), the set of linear transformations
thus obtained forms a representation. Denote the set of matrices representing
these linear transformations by R(G). R(G) is called the regular representa
tion, and any representation equivalent to R(G) is called a regular repre
sentation.
302 Selected Applications of Linear Algebra  VI
Let R(a) = [r i; (a)J. Then r i3 (a) = 1 if aa, = a l5 and otherwise r tf (a) = 0.
Since aa 3 = a j if and only if a = e, we have
S R (e) = g, (9.26)
S R (a) = for a^e. (9.27)
Thus,
2 /(a^S^a) = / (e^e) = gn r , (9.28)
a
so that by (9.21) a representation equivalent to D r (G) occurs n r times in
the regular representation.
Theorem 9.10. There are only finitely many inequivalent irreducible
representations.
proof. Every irreducible representation is equivalent to a component
of the regular representation. The regular representation is finite dimen
sional and has only a finite number of components. □
Furthermore,
 S R (^)S R (a) = S R (e)S R (e) = g 2 (9.29)
so that by (9.22)
I n r 8 = g. (9.30)
r
Let C i denote a conjugate class in G and h t the number of elements in
the class C € . Let m be the number of classes, and m' the number of in
equivalent representations. Since the characters are constant on con
jugate classes, they are really functions of the classes, and we can define
%i = X r ( a ) f° r any a e C t . With this notation formula (9.20)' takes the form
m
Ih<X< r X<' = g&rv (931)
Thus, the m' wtuples (\Jh 1 x 1 r , nK%£ , . . . , yjh m x m r ) are orthogonal and,
hence, linearly independent. This gives m' < m.
We can introduce a multiplication in V determined by the underlying
group relations. If a = 2f=i x i*i an< 3 ft = ^Li Vfii* we define
= 1( I Wi)**. (9.32)
This multiplication is associative and distributive with respect to the pre
viously detained addition. The unit element is e. Multiplication is not
commutative unless G is commutative.
9  Representations of Finite Groups by Matrices 303
Consider the elements
aeC,
y,b = J «b = b 2 b'ab = by, (9 33)
a.eC t aeCj.
and, hence, y^a = ay, for all a e V. Similarly, any sum of the y f commutes
with every element of V.
The importance of the y, is that the converse of the above statement is
also true. Any element in V which commutes with every element in V is a
linear combination of the y,. Let y = JLi c,a, be a vector such that yoc = ay
for every a e V. Then, in particular, yb = by, or b _1 yb = y for every beG.
If b _1 a,b is denoted by a,, we have
r = b 1 (ic l a i )b = ic l b 1 a i b
\ i=i / i=i
9 9
t=l 3=1
so that Ct = Cj. This means that all basis elements from a given conjugate
class must have the same coefficients. Thus, in fact,
m
7 = 2 c i7i
i=i
Since y^ also commutes with every element in V, we must have
m
yiyi=Ic) k y k . (9.34)
ft=0
Now, let Ci = 2aeCj ^ r ( a ) By exactly the same argument as before,
C* commutes with every matrix in D r (G). Thus, C/ must be a diagonal
matrix of the form
C t r = % r I r (9.35)
where I r is the identity of the rth representation. But >S(Q r ) = n r r)f and
S{Cf) = 2 a6Q ^(i)'(a)) = h iXi r . Thus,
/j.y r
^^ (936)
r "lXl n r i /rt T7\
^ = = — = 1, (9.37)
n r n r
where we agree that C x is the conjugate class containing the identity e. Also,
m
Q r C/ = l4c; (9.38)
304 Selected Applications of Linear Algebra \ VI
where these c) k are the same as those appearing in equation (9.34). This
means
m
a;=i
or
m
Vi% r = 24^ (939)
In view of equation 9.36, this becomes
n iXi "oXj _ y A "JcXk
n„ n„ fc=i n.
or
Kxl n al = n r i^khXk
k=l
to
= Xi r Ic)khkXk (940)
Thus,
m' m to'
2 Kx!^ai = 2 C 'A 2 **■#*"
r=l fc=l r=l
TO
= 2 c i k h k S R (a), where a g Q,
k=l
= cj lg , (9.41)
remembering that C x is the conjugate class containing the identity.
Suppose that C k contains the inverses of the elements of C t . Then x/ = Xk
Also, observe that y i y j contains the identity h t times if C, contains the inverses
of the elements of C t , and otherwise y^j does not contain the identity. Thus
c^ = h t ifj = k, and c) x = if j ^ k. Thus,
2%7x/ = f *«• (9  42)
r=l «j
Theorem 9.11. The number of inequivalent irreducible representations of a
finite group G is equal to the number of conjugate classes in G.
proof. With m the number of conjugate classes and m' the number of
inequivalent irreducible representations, we have already shown that m' < m.
Formula (9.42) shows that the w'tuples (%/, x*> • ■ • » xf) are mutually
orthogonal. Thus m < m' , and m = m . D
So far the only numbers that can be computed directly from the group
G are the c\. Formula (9.39) is the key to an effective method for computing
*il r
T
r\l
= [cJJ
T
V2
VrJ
T
9  Representations of Finite Groups by Matrices 305
all the relevant numbers. Formula (9.39) can be written as a matrix equation
in the form
(9.43)
where [c) k ] is a matrix with i fixed,;' the row index, and k the column index.
Thus, 77/ is an eigenvalue of the matrix [c) k ] and the vector (??/, r) 2 r , . . . , rj m r )
is an eigenvector for this eigenvalue. This eigenvector is uniquely determined
by the eigenvalue if and only if the eigenvalue is a simple solution of the
characteristic equation for [cjj. For the moment, suppose this is the case.
We have already noted that rj/ = 1. Thus, normalizing the eigenvector
so that rji = 1 will yield an eigenvector whose components are all the eigen
values associated with the rth representation.
The computational procedure: For a fixed i find the matrix [c) k ] and
compute its eigenvalues. Each of the m eigenvalues will correspond to one
of the irreducible representations. For each simple eigenvalue, find the
corresponding eigenvector and normalize it so that the first component is 1 .
From formulas (9.36) and (9.31) we have
m m% r _ v h ^xl
i=l h
= 1
(9.44)
i=i n.
This gives the dimension of each representation. Knowing this the characters
can be computed by means of the formula
%i
n r Y]i
(9.36)
Even if all the eigenvalues of [c* k ] are not simple, those that are may be
used in the manner outlined. This may yield enough information to enable
us to compute the remaining character values by means of orthogonality
relations (9.31) and (9.42). It may be necessary to compute the matrix
[ci k ] for another value of i. Those eigenvectors which have already been
obtained are also eigenvectors for this new matrix, and this will simplify
the process of finding the eigenvalues and eigenvectors for it.
Theorem 9.12. The dimension of an irreducible representation divides the
order of the group.
306 Selected Applications of Linear Algebra  VI
proof. Multiplying (9.39) by rj t r , we obtain
TO / TO \
fc=i \j)=i /
m i to \
= 2 244,W (945)
»=i\%=i /
Hence, rj/rj/ is an eigenvalue of the matrix [cj fc ] [c* p ]. If C t is taken to be
the class containing the inverses of the elements in C i5 we have
2f%V = lf^V
i=l «, i=l Hi
1 ™ — g 2
2 ^ <^» ^* == 2
(9.46)
£
Then — is an eigenvalue of the matrix 2™i t" [ C 3 J [ c iU All tne coefficients
of this matrix are integers. Hence, its characteristic polynomial has integral
coefficients and leading coefficient 1 . A rational solution of such an equation
must be an integer. Thus, — and, hence, — must be an integer. □
n/ n r
It is often convenient to summarize the information about the characters
in a table of the form :
D 1
D 2
K h 2 • • • h m
Q Q • • • C m
Xl X2 ' Xm
Xl X% ' Xm
v m v to ... v m
Ail a2 Am
(9.47)
The rows satisfy the orthogonality relation (9.31) and the columns satisfy
the orthogonality relation (9.42):
2 KXiXi = 8 d r
i=l
m
2 KxlXi = s d i
(9.31)
(9.42)
If some of the characters are known these relations are very helpful in
completing the table of characters.
Example. Consider the equilateral triangle shown in Fig. 8. Let (123)
denote the rotation of this figure through 120°; that is, the rotation maps P 1
9  Representations of Finite Groups by Matrices
Ps
Fig. 8
307
onto P 2 , P 2 onto P 3 , and P s onto P ± . Similarly, (132) denotes a rotation
through 240°. Let (12) denote the reflection that interchanges P 1 and P 2
and leaves P 3 fixed. Similarly, (13) interchanges P x and P 3 while (23) inter
changes P 2 and P 3 . These mappings are called symmetries of the geometric
figure. We define multiplication of these symmetries by applying first one
and then the other; that is, (123)(12) means to interchange P x and P 2 and
then rotate through 120°. We see that (123)(12) = (13). Including the iden
tity mapping as a symmetry, this defines a group G = {e, (123), (132), (12),
(13), (23)} of symmetries of the equilateral triangle.
The conjugate classes are C x = {e}, C 2 = {(123), (132)}, and C 3 = {(12),
(13), (23)}. It is easy to verify that
7272 = tyi + 72,
7273 = 273,
7s73 = 37i + 3y 2 .
Thus, we have [cfj is
"0
r
2
3
3 o_
The eigenvalues are 0, 3, and —3. Taking the eigenvalue i?, 1 = 3, we get the
eigenvector (1 , 2, 3). From (9.44), we get n x = 1. For the eigenvalue rj 3 2 =
—3 we get the eigenvector (1 ,2, 3) and n 2 = 1. For the eigenvalue r? 3 3 = 0,
we get the eigenvector (1,1,0) and « 3 = 2. Computing the characters
by means of (9.36)', we get the character table
1 2 3
Cj C 2 I3
D 1
1
1
1
D 2
1
1
1
D*
2
1
308
Selected Applications of Linear Algebra  VI
The dimensions of the various representations are in the first column, the
characters of the identity. The most interesting of the three possible ir
reducible representations is D 3 since it is the only 2dimensional representa
tion. Since the others are 1 dimensional, the characters are the elements
of the corresponding matrices. Among many possibilities we can take
D 3 (e) =
Z) 3 ((12)) =
"1
0"
_0
1_
"0
r
1
Z> 3 ((123)) =
£> 3 ((13)) =
"0
r
1
i_

1
0"
1
i_
Z) 3 ((132)) =
D*((23)) =
1
0"
_
1
1
"1
r
i_
BIBLIOGRAPHICAL NOTES
The necessary background material in group theory is easily available in G. Birkhoff
and S. MacLane, A Survey of Modern Algebra, Third Edition, or B. L. van der Waerden,
Modern Algebra, Vol. 1. More information on representation theory is available in V. I.
Smirnov, Linear Algebra and Group Theory, and B. L. van der Waerden, Gruppen von
Linearen Transformationen. F. D. Murnaghan, The Theory of Group Representations,
is encyclopedic.
EXERCISES
The notation of the example given above is particularly convenient for repre
senting permutations. The symbol (123) is used to represent the permutation,
"1 goes to 2, 2 goes to 3, and 3 goes to 1." Notice that the elements appearing
in a sequence enclosed by parentheses are cyclically permuted. The symbol
(123)(45) means that the elements of {1, 2, 3} are permuted cyclically, and the
elements of {4, 5} are independently permuted cyclically (interchanged in this
case). Elements that do not appear are left fixed.
1. Write out the full multiplication table for the group G given in the example
above. Is this group commutative?
2. Verify that the set of all permutations described in Chapter IIIl form a group.
Write the permutation given as illustration in the notation of this section and verify
the laws of combination given. The group of all permutations of a finite set
S = {1, 2, ...,«} is called the symmetric group on n objects and is denoted by 6„.
Show that S n is of order n !. A subgroup of <3 n is called a. group of symmetries , or a
permutation group.
3. Show that the subset of <3 n consisting of even permutations forms a subgroup
of <B n . This subgroup is called the alternating group and is denoted by 9t n . Show
that 5l n is a normal subgroup.
4. For any group G and any a e G, let Z> x (a) be the 1 x 1 unit matrix, D x (a) =
[1]. Show that D 1 (G) = {D x (si)  a G G} is a representation of G. This representa
tion is called the identity representation.
9  Representations of Finite Groups by Matrices 309
5. S 3 be the symmetric group described in the example given above. We showed
there that S 3 has three inequivalent irreducible representations. One of them is
the identity representation ; another is the 2 x 2 representation which we described.
Find the third one.
6. Show that any 1 dimensional representation of a finite group is always in
unitary form.
7. Give the 2 x 2 irreducible representation of <5 3 in unitary form.
8. Show that a finite group G is commutative if and only if every irreducible
representation is of dimension 1 .
9. Show that a finite commutative group of order n has n inequivalent irreducible
representations.
10. Let G be a cyclic group of order n. Find the n irreducible inequivalent
representations of G.
1 1 . There are two nonisomorphic groups of order 4. One is cyclic. The other
is of the form 23 = {e, a, b, c} where a 2 = b 2 = c 2 = e and ab = c, ac = b,
be = a. 23 is called the fourgroup. Find the four inequivalent irreducible repre
sentations for each of these groups.
12. Show that if G is a group of order/? 2 , where/? is a prime number, then G is
commutative.
13. Show that all groups of orders, 2, 3, 4, 5, 7, 9, 11, and 13 are commutative.
14. Show that there is just one commutative group for each of the orders 6,
10, 14, 15.
15. Show that a noncommutative group of order 6 must have three irreducible
representations, two of dimension 1 and one of dimension 2. Show that this
information and the knowledge that one of the representations must be the identity
representation determines five of the nine numbers that appear in the character
table. How many conjugate classes can there be? What are their orders? Show
that we now know enough to determine the remaining elements of the character
table. Show that this information determines the group up to an isomorphism;
that is, any two noncommutative groups of order 6 must be isomorphic.
16. Show that if every element of a group is of order 1 or 2, then the group
must be commutative.
17. There are five groups of order 8 ; that is, every group of order 8 is isomorphic
to one of them. Three of them are commutative and two are noncommutative.
Of the three commutative groups, one contains an element of order 8, one contains
an element of order 4 and no element of higher order, and one contains elements
of order 2 and no higher order. Write down full multiplication tables for these
three groups. Determine the associated character tables.
18. There are two noncommutative groups of order 8. One of them is generated
by the elements {a, b} subject to the relations, a is of order 4, b is of order 2,
and ab = ba 3 . Write out the full multiplication table and determine the associated
character table. An example of this group can be obtained by considering the
group of symmetries of a square. If the four corners of this square are numbered,
310 Selected Applications of Linear Algebra  VI
a representation of this group as a permutation group can be obtained. The other
group of order 8 is generated by {a, b, c} where each is of order 4, ab = c, be = a,
ca = b, and a 2 = b 2 = c 2 . Show that ab = b 3 a = ba 3 . Write out the full multi
plication table for this group and determine the associated character table. Com
pare the character tables for these two nonisomorphic groups of order 8.
The above exercises have given us a reservoir of representations of groups of
relatively small order. There are several techniques for using these representations
to find representations of groups of higher order. The following exercises illustrate
some of these techniques.
19. Let G x be a group which is the homomorphic image of the group G a . Let
Z>(G X ) be a representation of G v Define a homomorphism of G 2 onto D(G^) and
show that D(G^) is also a representation of G 2 .
20. Consider the two noncommutative groups of order 8 given in Exercise 18.
Show that H = {e, a 2 } is a normal subgroup (using the appropriate interpretation
of the symbol "a" in each case). Show that, in either case, GjH is isomorphic
to the fourgroup 23. Show how we can use this infromation to obtain the four
1 dimensional representations for each of these groups. Show how the characters
for the remaining representation can be obtained by use of the orthogonality
relations (9.31) and (9.42).
21. In a commutative group every subgroup is a normal subgroup. In Exercise
10 we determined the character tables for a cyclic group. Using this information
and the technique of Exercise 19, find the character tables for the three commutative
groups of order 8.
22. Show that S n has a 1 dimensional representation in which every element
of <H TO is mapped onto [1] and every element not in <JI n is mapped onto [1].
23. Show that if D r (G) is a representation of G of dimension n, where D r (a) =
[flf/a)], and D S (G) is a representation of dimension m, where Z) s (a) = [a^(a)],
then Z) rX '(Q, where D rX, (a) = K fc; ?j(a) = fl^(a)a* 4 (a)], is a representation of G
of dimension mn. D rXS (G) is known as the Kronecker product of D r (G) and D S (G).
24. Let S r (a) be the trace of a for the representation D r (G), S s (a) the trace for
D S (G), and S"X"(a) the trace for D r *'(G). Show that S r * s (a) = 5 r (a)5 s (a).
25. The commutative group of order 8, with no element of order higher than 2,
has the following three rows in the associated character table:
11 1 111 11
1 11 11 111
1 1 11 111 1.
Find the remaining five rows of the character table.
26. The commutative group of order 8 with an element of order 4 but no element
of order higher than 4 has the following two rows in the associated character table:
1 1 1 11111
1 i 1 i 1 / 1 i.
Find the remaining six rows of the character table.
9  Representations of Finite Groups by Matrices 311
27. Let 77 and a be permutations of the set S ={1,2,..., n), and let a = tT x ott
be conjugate to a. Show that if a'(i) =j, then a^ii)) = tt(j). Let a' be repre
sented in the notation of the above example in the form o' = (■■• ij ■••)••• .
Show that a is represented in the form a = ( • • • rr{i)it(j) •■)■••. As an example,
let a' = (123)(45) and ■* = (1432). Compute a = W 'it 1 directly, and also replace
each element in (123)(45) by its image under n.
28. Use Exercise 27 to show that two elements of S n are conjugate if and only
if their cyclic representations have the same form; for example, (123)(45) and
(253)(14) are conjugate. (Warning: This is not true in a permutation group, a
subgroup of a symmetric group.) Show that <5 4 has five conjugate classes.
29. Use Exercise 28, Theorems 9.11 and 9.12, and formula (9.30) to determine
the dimensions of the irreducible representations of S 4 .
30. Show that three of the conjugate classes of S 4 fill out <H 4 .
31. Use Exercises 22 and 30 to determine the characters for the two 1 dimen
sional representations of S 4 . Use Exercise 29 to determine one column of the
character table for S 4 , the column of the conjugate class containing the identity
element.
32. Show that 33 = {e, (12)(34), (13)(24), (14)(23)} is a subgroup of S 4 iso
morphic to the fourgroup. Show that 93 is a normal subgroup of S 4 . Show that
each coset of 93 contains one and only one of the elements of the set {e, (123),
(132), (12), (13), (23)}. Show that the factor group S 4 /93 is isomorphic to S 3 .
33. Use Exercises 19 and 32 and the example preceding this set of exercises to
determine a 2dimensional representation of <S 4 . Determine the character values
of this representation.
34. To fix the notation, let us now assume that we have obtained part of the
character table for S 4 in the form:
1
6
8
6
3
Cx
c 2
c 3
Q
c.
D 1
1
1
1
1
1
D*
1
1
1
1
1
D 3
2
1
2
D i
3
Z> 5
3
Show that if Z) 4 is a representation of S 4 , then the matrices obtained by multi
plying the matrices in D 4 by the matrices in D 2 is also a representation of S 4 .
Show that this new representation is also irreducible. Show that this representation
must be different from D 4 , unless D 4 has zero characters for C 2 and C 4 .
35. Let the characters of D 4 be denoted by
Z> 4 3 a b c d.
312 Selected Applications of Linear Algebra  VI
Show that
3 + 6a + 8b + 6c + 3 d =
3  6a + 8b  6c + 3d =
6  8b + 6d = 0.
Determine b and d and show that a = — c. Show that a 2 = 1. Obtain the complete
character table for 6 4 . Verify the orthogonality relations (9.31) and (9.42) for this
table.
10 I Application of Representation Theory to Symmetric Mechanical Systems
This section depends directly on the material in the previous two sections,
8 and 9.
Consider a mechanical system which is symmetric when in an equilibrium
position. For example, the • ozone molecule consisting of three oxygen
atoms at the corners of an equilateral triangle is very symmetric (Fig. 9).
This particular system can be moved into many new positions in which it
looks and behaves the same as it did before it was moved. For example,
the system can be rotated through an angle of 120° about an axis through
the centroid perpendicular to the plane of the triangle. It can also be reflected
in a plane containing an altitude of the triangle and perpendicular to the
plane of the triangle. And it can be reflected in the plane containing the
triangle. Such a motion is called a symmetry of the system. The system
above has twelve symmetries (including the identity symmetry, which is to
leave the system fixed).
Since any sequence of symmetries must result in a symmetry, the sym
metries form a group G under successive application of the symmetries as the
law of combination.
Let X = (x 1} . . . , x n ) be an wtuple representing a displacement of the
system. Let a be a symmetry of the system. The symmetry will move the
Fig. 9
10 I Application of Representation Theory to Symmetric Mechanical Systems 313
system to a new configuration in which the displacement is represented by
X'. The mapping of X onto X' will be represented by a matrix D(a); that
is, D(a)X = X'. If a new symmetry b is now applied, the system will be
moved to another configuration represented by X" where X" = D(b)X'.
But since ba moves the system to the configuration X" in one step, we have
X" = D(ba)X = D(b)D(a)X. This holds for any X so we have £>(ba) =
D(b)D(a). Thus, the set D(G) of matrices obtained in this way is a repre
sentation of the group of symmetries.
The idea behind the application of representation theory to the analysis
of symmetric mechanical systems is that the irreducible invariant subspaces
under the representation D(G) are closely related to the principal axes of
the system.
Suppose that a group G is represented by a group of linear transformations
{<r a } on a vector space V. Let /be a Hermitian form and let g be the sym
metrization of/ defined by
g(",£) = 2/(tfa(a),<Ta(£)). ( 1(U )
g a
Let A = {. . . , a/, . . . , a. r nr . . .} be a basis for V such that {a/, . . . , a. r n )
is a basis for the irreducible subspace on which G is represented by D r (G)
in unitary form; that is,
n r
tfa(a/)=2<aa)< ( 10  2 )
where D r (a) = [a r H (a)] is unitary. Then, by (9.17)',
g a
= lf
g a
3=1 fc=l
= 22 I«Ii(a»)/(a>/)
g a 3=1 fc=i
laWaKiM /«.«*')
= 2 2
= 2 2MA./K.O d° 3 )
If there is at most one invariant subspace corresponding to each irreducible
representation of G, the matrix representing g with respect to the basis A
would be a diagonal matrix. If a given irreducible representations occurs
more than once as a component of D(G), then terms off the main diagonal
can occur, but their appearance depends on the values of /(a/, a fc s ). If/ is
314 Selected Applications of Linear Algebra  VI
left invariant under the group of transformations — that is,/(cr a (a), <r a )(/?)) =
/(a, /?) for all a e G — then g =/and the same remarks apply to/.
By a symmetry of a mechanical system we mean a motion which preserves
the mechanical properties of the system as well as the geometric properties.
This means that the quadratic forms representing the potential energy and
the kinetic energy must be left invariant under the group of symmetries.
If a coordinate system is chosen in which the representation is unitary and
decomposed into its irreducible components, considerable progress will be
made toward finding the principal axes of the system. If D(G) contain each
irreducible representation at most once, the decomposition of the repre
sentation will yield the principal axes of the system. If the system is not very
symmetric, the group of symmetries will be small, there will be few in
equivalent irreducible representations, and it is likely that the reduced form
will fall short of yielding the principal axes. (As an extreme case, consider
the situation where the system has no symmetries except the identity.) How
ever, in that part of the representing matrices where r ^ s the terms will be
zero. The problem, then, is to find effective methods for finding the basis A
which reduces the representation.
The first step is to find the irreducible representations contained in D{G).
This is achieved by determining the trace S D for D and using formula (9.21)'.
The trace is not difficult to determine. Let C/(a) be the number of particles
in the system left fixed by the symmetry a. Only coordinates attached to
fixed particles can contribute to the trace S D (a). If a local coordinate system
is chosen at each particle so that corresponding axes are parallel, they will
be parallel after the symmetry is applied. Thus, each local coordinate system
(at a fixed point) undergoes the same transformation. If the local coordinate
system is Euclidean, the local effect of the symmetry must be represented
by a 3 x 3 orthogonal matrix since the symmetry is distance preserving.
The trace of a matrix is the sum of its eigenvalues, since that is the case when
it is in diagonal form. The eigenvalues of an orthogonal matrix are of
absolute value 1. Since the matrix is real, at least one must be real and the
others real or a pair of complex conjugate numbers. Thus, the local trace is
±1 + e " + e ~ ie = ±1 + 2 cos 6,
and (10.4)
S D (*)= t/(a)(±l + 2cos0).
The angle d is the angle of rotation about some axis and it is easily determined
from the geometric description of the symmetry. The +1 occurs if the
symmetry is a local rotation, and the —1 occurs if a mirror reflection is
present.
Once it is determined that the representation D r (G) is contained in D(G),
the problem is to find a basis for the corresponding invariant subspace.
10 I Application of Representation Theory to Symmetric Mechanical Systems 315
If {a/, . . . , a^} is the required basis, we must have
«)=IaUa)<. ( 10  5 )
3=1
If a/ is represented by X t (unknown) in the given coordinate system, then
cr a (a/") is represented by D(n)Xi. Thus, we must solve the equations
D(*)Xi = 2 a r MX„ i = 1, . . . , n„ (10.6)
3=1
simultaneously for all a e G. The a r H (a) can be computed once for all and
are presumed known. Since each X t has n coordinates, there are n • n r
unknowns. Each matric equation of the form (10.6) involves n linear equa
tions. Thus, there are g • n • n r equations. Most of the equations are re
dundant, but the existence or nonexistence of the rth representation in
D{G) is what determines the solvability of the system. Even when many
equations are eliminated as redundant, the system of linear equations to be
solved is still very large. However, the system is linear and the solution can
be worked out.
There are ways the work can be reduced considerably. Some principal
axes are obvious for one reason or another. Suppose that Y is an ntuple
representing a known principal axis. Then any other principal axis repre
sented by X must satisfy the condition
Y T BX = (10.7)
since the principal axes are orthogonal with respect to the quadratic form B.
There is also the possibility of using irreducible representations in other
than unitary form in the equations of (10.6). The basis obtained will not
necessarily reduce A and B to diagonal form. But if the representation uses
matrices with integral coefficients, the computation is sometimes easier,
and the change to an orthonormal basis can be made in each invariant
subspace separately.
BIBLIOGRAPHICAL NOTES
There are a number of good treatments of different aspects of the applications of group
theory to physical problems. None is easy because of the degree of sophistication required
for both the physics and the mathematics. Recommended are: B. Higman, Applied
GroupTheoretic and Matrix Methods; J. S. Lomont, Applications of Finite Groups.
T. Venkatarayudu, Applications of Group Theory to Physical Problems; H. Weyl, Theory
of Groups and Quantum Mechanics; E. P. Wigner, Group Theory and Its Application to the
Quantum Mechanics of Atomic Spectra.
316 Selected Applications of Linear Algebra  VI
EXERCISES
The following exercises all pertain to the ozone molecule described at the
beginning of this section. However, to reduce the complexity of analyzing this
system we make a simplification of the problem. As described at the beginning
of this section, the phase space for the system is of dimension 9 and the group of
symmetries is of order 12. This system has already been discussed in the exercises
of Section 8. There we assumed that the displacements in a direction perpendicular
to the plane of the triangle could be neglected. This has the effect of reducing the
dimension of the phase space to 6 and the order of the group of symmetries to 6.
This greatly simplifies the problem without discarding essential information,
information about the vibrations of the system.
1. Show that if the ozone molecule is considered as embedded in a 2dimensional
space, the group of symmetries is of order 6 (instead of 12 when it is considered as
embedded in a 3dimensional space). Show that this group is isomorphic to 6 3 ,
the symmetric group on three objects.
2. Let (x lf y x , x 2 , y 2 , x z , y z ) denote the coordinates of the phase space as illus
trated in Fig. 6 of Section 8. Let (12) denote the symmetry of the figure in which
P x and P 2 are interchanged. Let (123) denote the symmetry of the figure which
corresponds to a counterclockwise rotation through 120°. Find the matrices
representing the permutations (12) and (123) in this coordinate system. Find
all matrices representing the group of symmetries. Call this representation
D{G).
3. Find the traces of the matrices in D(G) as given in Exercise 2. Determine
which irreducible representations (as given in the example of Section 9) are con
tained in this representation.
4. Show that since D(G) contains the identity representation, 1 is in eigenvalue
of every matrix in D{G). Show that the corresponding eigenvector is the same for
every matrix in D{G). Show that this eigenvector spans the irreducible invariant
subspace corresponding to the identity representation. On a drawing like Fig. 7,
draw the displacement corresponding to this eigenvector. Give a physical inter
pretation of this displacement.
5. There is one other 1dimensional representation in D(G). Show that for a
1 dimensional representation a character value is also an eigenvalue. Determine
the vector spanning the irreducible invariant subspace corresponding to this
representation. Draw the displacement represented by this eigenvector and give
a physical interpretation of this displacement.
6. There are two 2dimensional representations. One of them always corre
sponds to an invariant subspace that always appears in this type of problem. It
corresponds to a translation of the molecule in the plane containing the molecule.
Such a translation does not distort the molecule and does not play a role in deter
mining the vibrations of the molecule. Find a basis for the irreducible invariant
subspace corresponding to this representation.
10 I Application of Representation Theory to Symmetric Mechanical Systems 317
7. In the previous exercises we have determined two 1 dimensional subspaces
and one 2dimensional subspace of the representation space for D(G). There
remains one more 2dimensional representation to determine. At this stage the
easiest thing to do is to use the orthogonality relations in formula (10.7) to find a
basis for the remaining irreducible invariant subspace. Find this subspace.
8. Consider the displacement vectors { 5 = (0, 1, ^3/2, J,  V3/2, ),
£ 6 = (  V3/2, i, 0, 1 , V3/2, !)}. Show that they span an irreducible invariant
subspace of the phase space under D(G). Draw these displacements on a figure
like Fig. 7. Similarly, draw (£ 5 + l 6 ). Interpret these displacements in terms
of distortions of the molecule and describe the type of vibration that would result
if the molecule were started from rest in one of these positions.
Note: In working through the exercises given above we should have seen that
one of the 1 dimensional representations corresponds to a rotation of the molecule
without distortion, so that no energy is stored in the stresses of the system. Also,
one 2dimensional representation corresponds to translations which also do not
distort the molecule. If we had started with the original 9dimensional phase space,
we would have found that six of the dimensions correspond to displacements
that do not distort the molecule. Three dimensions correspond to translations in
three independent directions, and three correspond to rotations about three
independent axes. Restricting our attention to the plane resulted only in removing
three of these distortionless displacements from consideration. The remaining
three dimensions correspond to displacements which do distort the molecule,
and hence these result in vibrations of the system. The 1 dimensional representation
corresponds to a radial expansion and contraction of the system. The 2dimensional
representation corresponds to a type of distortion in which the molecule is expanded
in one direction and contracted in a perpendicular direction.
Appendix
A collection of
matrices with
integral elements
and inverses with
integral elements
Any numerical exercise involving matrices can be converted to an equivalent
exercise with different matrices by a change of coordinates. For example,
the linear problem
AX = B (A.1)
is equivalent to the linear problem
A'X = B'
(A.2)
where A' = PA and B' = PB and P is nonsingular. The problem (A.2) even
has the same solution. The problem
A"Y=B (A.3)
where A" = AP has Y = P _1 Zas a solution if X is a solution of (A.l).
It should be clear enough how these modifications can be combined.
Other exercises can be modified in a similar way. For this purpose it is most
convenient to choose a matrix P that has integral elements (integral matrices).
For (A.3), it is also desirable to require P _1 to have integral elements.
Let P be any nonsingular matrix, and let D be any diagonal matrix of the
same order. Compute A = PDP 1 . Then D = P~ X AP. A is a matrix similar
to the diagonal matrix D. The eigenvalues of A are the elements in the main
diagonal of D, and the eigenvectors of A are the columns of P. Thus, by
choosing D and P appropriately, we can find a matrix A with prescribed
eigenvalues (whatever we choose to enter in the main diagonal of D) and
prescribed eigenvectors (whatever we put in the columns of P). If P is orthog
onal, A will be orthogonal similar to a diagonal matrix. If P is unitary, A
will be unitary similar to a diagonal matrix.
It is extremely easy to obtain an infinite number of integral matrices with
integral inverses. Any product of integral elementary matrices will be integral.
319
320 Matrices and Inverses with Integral Elements  Appendix
If these elementary matrices have integral inverses, the product will have an
integral inverse. Any elementary matrix of Type III is integral and has
an integral inverse. If an elementary matrix of Type II is integral, its inverse
will be integral. An integral elementary matrix of Type I does not have an
integral inverse unless it corresponds to the elementary operation of multiply
ing by ±1. These two possibilities are rather uninteresting.
Computing the product of elementary matrices is most easily carried out
by starting with the unit matrix and performing the corresponding elementary
operation in the right order. Thus, we avoid operations of Type I, use only
integral multiples in operations of Type II, and use operations of Type III
without restriction.
For convenience, a short list of integral matrices with integral inverses is
given. In this list, the pair of matrices in a line are inverses of each other
except that the inverse of an orthogonal, or unitary, matrix is not given.
P
P
i
"2
r
" 3
r
5
3_
5
2_
"5
8"
' 5
8"
3
5_
3
5_
"3 
8"
"11
8"
4 
11_
_ 4
3_
3
8"
" 11
8"
4
11_
4
3_
"4 3
2"
"1
1 4"
3 5
2
1
2
_2 2
1
4
2 11_
"2 5
5"
" 43
5 25"
2 3
8
10
1 6
_3 8
7_
_7
1 4 _
P
px
10 6
3"
" 1
3"
8 3
2
2
1 4
3
2
1_
_7
2
!8_
Appendix  Matrices and Inverses with Integral Elements
321
P
pi
"2
5
5"
" 43
5 
25
2
3
8
10
1
6
3
8
7_
7
1
4
"2
5
8"
"73 43
25"
4
5
13
17 10
6
1
6
—
1_
29 17
10.
"1
2
3"
"2
r
2
3
4
3 
2
3
4
6_
1
2
1_
" 4 3
2
0"
' 2
1
1 0"
5 4
3
1

2
2 2
1
2
2
1
_ 11 6
4
1_
8
3 
3 1_
" 2
1
0"
' 1
1
1"
1
1
1
2 
2
1
1_
1
2
1_
" 2
1
0"
"1
i r
1
1
1 .
2 2
1
1_
1
2 1.
" 4 3
2
r
" :
2 1
1
5 4
3
i
7 3
1 1
2 2
1
—
i
10 5
2 1
_ 11 6
4
3_
8 3
3 1
Orthogor
lal
1
"3
_z
r
5
.4
J_
"1
2
2"
1
3
2 
2
1
2
1
2
322
Matrices and Inverses with Integral Elements  Appendix
Orthogonal
"2 1 2"
1
27
1 2 2
2 2 1.
12 2"
2 1 2
22 1_
_7 _4 _4"
4 1 8
4 i
23
10
10
1
10
1
87
10 2
17
25 2
25
56
56
"2
56
56"
49
32
32
49_
1
33
6
9
17
4
28
6
7
6
4
32
7
28"
7
16
1
26
Unitary
"4 3/~
i 1
■ 7/ 7
7 + 17i 17 + li '
17 + 7/ 7  17/
1
"4 3i"
5
_3» 4 .
1
"7 + /  1 + ir
10
_i +
7/ 7
— i
Appendix  Matrices and Inverses with Integral Elements
Unitary
"cos 6 i sin (
i sin cos 6 _
1 ["(1 + i)e~ ie (1  i)e ie ~
2 _(i + i)e i0 (1  i)e i0 _
323
(0 real)
Answers
to selected
exercises
11
1^. If / and g are (continuous, integrable, differentiable m times, satisfy the
differential equation) and a is a constant, then f + g and af also are con
tinuous, etc.). With the exception of A\ and B\, any vector space axiom
which is satisfied in a set is satisfied in any subset.
5. B\ is not satisfied if a is negative.
6. a = (a x a)a = a'^aoC) = a^O =0.
9 . („)(_i, 2, 1,0); » (5, 8, 1,2); (c) (6, 15,0,3); (rf)(5, 1,3,
1).
12
1. 3/>! + 2/» 8  5p 3 + 4/? 4 = 0.
2. The set of all polynomials of degree 2 or less, and the zero polynomial.
3. Every subset of three polynomials is maximal linearly independent.
6. No polynomial is a linear combination of the preceding ones.
7. 1 cannot be expressed as a linear combination of polynomials divisible by x  1 .
8. (a) dependent; (b) dependent; (c) independent.
13
3. {(1, 2, 1, 1), (0, 1, 2, 1), (1, 1, 0, 0), (0, 0, 1, 1)}.
4. {(1, 2, 3, 4), (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0)}, for example.
14
1. (a), (b), and (d) are subspaces. (c) and (e) do not satisfy A\. ..„.,
3. The equation used in (c) is homogeneous. This condition is essential if A\
and B\ are to be satisfied. Why?
325
326 Answers to Selected Exercises
5. (2, 1, 3, 3) = (1, 1, 0, 0) + 3(1, 0, 1, 1), (0, 1, 1, 1) = (1, 1, 0, 0) 
(1, 0, 1, 0); (1, 1, 0, 0) = (2, 1, 3, 3) + f(0, 1, 1, 1), (1, 0, 1, 1) =
(2 —1 3 3) + (0 1 —1, —1).
6. f(l,' 1, l' l', 1), (l! 0,' 1,' 0, 1), (0, 1, 1, 1, 0), (2, 0, 0, 1, 1)} is a basis for VV.
8. W x nW 2 = <(l, 2, 1, 2)>, W 1 = <(l, 2, 1, 2), (1, 2, 3, 6)>, W 2 = <(l,
2, 1,2), (1, 1, 1, 1)>, W x + W 2 = <(l, 2, 1, 2), (1, 2, 3, 6), (1, 1, 1, 1)>.
9. W x n W 2 is the set of all polynomials divisible by (x — l)(x — 2). W 1 + VV 2 =
P.
11. If W x £ W 2 and W 2 £ W l5 there exist c^e Wj  VV 2 and <x 2 e W 2  W x .
a i + a 2 ^ W x since a x e VV X and a 2 ^ W x . Similarly, a. x + a 2 ^ W 2 . Since
x i + a 2^iU W 2 , VVj^ (j W 2 is not a subspace.
12. {(1, 1, 1, 0), (2, 1, 0, 1)} is a basis for the subspace of solutions.
16. Since W 1 <= VV, W 1 + (VV n VV 2 ) c VV. Every a 6 W n (W x + VV 2 ) can be
written in the form a = a x + a 2 where a x e W x and <x 2 e VV 2 . Since W x <= VV,
a x e VV. Thus a 2 e VV and a E (VV n W^ + (VV n VV 2 ) = W x + (VV n W 2 ).
Thus VV = VV n (W x + W 2 ) = Wj + (VV n VV 2 ). Finally, it is easily seen
that this last sum is direct.
IIl
2. {o x + ffgXOa* X 2» = ( x l + X 2, ~ X 1  X 2)> ^l^CO^l. ^2)) = (^2. _a; l)»
ffa^CCa?!, » 2 )) = (a^i)
4. The kernel of a is the set of solutions of the equations given in Exercise 12 of
Section 4, Chapter I.
5. {(1, 0, 1), (0, 1, 2)} is a basis of a(U). {(4, 7, 5)} is a basis of K(a).
7. See Exercise 2.
12. By Theorem 1.6, P (<x) = dim V = dim {K(a) n V'} + dim t(V') < dim K{r) +
dim tg{U) = v{t) + p(tct).
13. By Exercise 12, p (to) > P (a)  v( T ) = p(ff)  (m  p(r)) = p(<r) + />(t) 
w. The other part of the inequality is Theorem 1.12.
14. By Exercise 13, v(t<j) = n — p(to) < n — (p(cr) 4 p(t) — m) = (n — p(a)) +
(m — p(r)) = v(o) + v{t). v{ra) < n since K(tg) <= L/. The inequality
v(o) < v(to) follows from the fact that K(a) <= K(ra). Since t<t(L/) e
t(V) we have p(ror) < p (t). Thus v(t<t) = n — p(to) > n — p(r) = n — m +
Kt).
15. By Exercise 14, v(to) = v(a).
17. By Exercise 14, v{tg) = i>(t).
18. p(o x + <r 2 ) = dim {a x + o 2 )(U) < dim {^(U) + a 2 (U)} = dim ^(U) + dim or^U) 
dim (0 X (U) n 2 (U)} < P {a x ) + P {a 2 ).
19. Since p(a 2 ) = p{ — a 2 ), Exercise 18 also says that p{a x — a 2 ) <; P {a x ) + p{o 2 ).
Then P {a x ) = P (a 2  (a x + a 2 )) < P {o 2 ) + P {o 1 + a 2 ). By symmetry, P (o 2 ) <
p(°i) + p( ff i + <T 2)
II2
r 37"
2.
Answers to Selected Exercises
327
3. AB =
_4 _4 _4 4"
4 4 4 4
0J,
BA =
4
12
4
12
6
14
6
14
5
13
5
13
5"
13
5
13
Notice that AB and BA have different ranks.
" 3 1"
_l 2 
5.
" 3 r
_5 2_
6.
r 2 1_ i
5 5
1 3
L5 5J
(«)
"I 0"
_ 1.
(6)
 o r
j
"1 0"
1.
(c)
" 5 _1
2 2
1 5
. 2 2J
J
"2
_0
(«0
"i r
(*
)
5 121
13 13
12 5
13 13J
(/)
r2
3
J
in
3
1
3_
,
"1
_0
0"
0_
a
9. (a) Reflection in the line y = x. See 8(6). (6) Projection onto %axis parallel
to x 2 axis, followed by 90° rotation counterclockwise. See 8(/) and 8(e).
(c) Projection onto a^axis parallel to line x 1 + x 2 = 0. See 8(/). (d) Shear
parallel to x 2 axis. See $(d). (e) Stretch » x axis by factor b, stretch » 2 axis by
factor c. See 8(c). (/) Rotation counterclockwise through acute angle
6 = arccos f .
10. (a) y =x (onto, fixed), (6) none, (c)x 2 = (onto, fixed), x t + x 2 = (into),
(</) Xl = (onto, fixed), (e) x 2 = (onto, stretched by factor b), x 1 = (onto,
stretched by factor a), (/) none.
1 1
11.
1
1 1
12. If {d x , d 2 ,..., d n } is the set of elements in the diagonal of D, then multiplying
A by D on the right multiplies column k by d k and multiplying A by D on the
left multiplies row k by d k .
13. A is a diagonal matrix.
Fa b~
15. Let /be represented by
16. The linear transformation, and thereby the function /(a; + yi), can be repre
~a K
sented by the matrix
17. For example, A =
b
1 o
B
n
II3
2. A 2 = I.
3. (0, 0, 0). See Theorem 3.3.
5
3.
5J'
6.
o r
1 o
"1 1
1
Answers to Selected Exercises
1 0"
1
1
328
o o r
10
.010.
9. If a is an automorphism and S is a subspace, o(S) is of dimension less than or
equal to dim S. Since K(a) is of dimension zero in V, it is of dimension zero
when confined to S. Thus dim o(S) = dim S.
10. For example, A =
ri on
"1 o o
1
B =
1
Lo oj
II4
1
2
1
1
1
1
1
1
1
5. P =
"0
1
r
1
l
_1
1
o_
pi =2
"
i
V31
3.
2
V3
2
1
L 2
2 .
1
1
r
1
1
l
1
1
l
6. PQ is the matrix of transition from A to C. The order of multiplication is
reversed in the two cases.
II5
1. (a), (c), {d).
2. (a) 3, (b) 3, (c) 3, (d) 3, (<?) 4.
3. Both A and 5 are in Hermite normal form. Since this form is unique under
possible changes in basis in R 2 , there is no matrix Q such that B = Q~ 1 A.
II6
1. (a) 2, (b) 2, (c) 3.
2. (a) Subtract twice row 1 from row 2. (b) Interchange row 1 and row 3.
(c) Multiply row 2 by 2.
3. (From right to left.) Add row 1 to row 2, subtract row 2 from row 1, add
row 1 to row 2, multiply row 1 by — 1 . The result interchanges row 1 and row 2.
"1
5. (a)
6. (a)
(b)
(b)
1
1 2
3
2
1
2
1
Answers to Selected Exercises ^^
II7
1. For every value of x z and » 4 , (x lt x 2 , x z , z 4 ) = x z (\, 1,1,0)+ a? 4 (2, 1, 0, 1) is
a solution of the system of equations. Since the system of homogeneous
equations is of rank 2 and these two solutions are linearly independent, they
span the space of solutions.
2. <(l,2, 1,0>. t .
3. (a) (2, 3, 0, 0) + <(1, 2, 1, 0), (1, 1, 0, 1)>, (b) no solution.
4. (3, 5, 0) + <(5, 7, 1)>.
5. (3,2,0,0,1) + ((3,0,1,0,0), (1,2,0, 1,0)>.
6. If the system has m equations and n unknowns, the augmented matrix is
m x (n + 1). The reduced form of the augmented matrix contains the reduced
form of the coefficient matrix in the first n columns. Their ranks are equal
if and only if the last column of the reduced form of the augmented matrix does
not start a nonzero row.
II8
2. {(1, 0, 0, 0, 1), (0, 1, 0, 0, 1), (0, 0, 1, 0, 0), (0, 0, 0, 1, 1)}.
3. {(1, 0, 2, f), (0, 1, 0, f)} is a standard basis of the subspace spanned by the
first set and {(1 ,0,1, 0), (0, 1 , 0, f )} is a standard basis for the second. Hence
these subspaces are not identical.
4. {(1,0, 0, 1), (0, 1,0, f), (0, 0, 1, i)} is a standard basis of W 1 + VV 2 .
5. {(1, 0, 1,0, 2), (0, 1, 2, 0, 1), (0, 0, 0, 1, 1)} is a stand basis. x x  2x 2 +
x s = 0, 2^!  x 2  x 4 + x 5 = is a characterizing system.
6. W x = <(£, 0, h 1), (1, 1,0, 0)>, £! = <(!, 1,0, ), (0,0, 1, £)>, W 2 =
<(2, 3, 0, 1), (3, 4, 1, 0)>, E 2 = <(1, 0, 3, 2), (0, 1, 4, 3)>, E x + £ 2 =
<(1, 0, 0, ), (0, 1,0, 1), (0, 0, 1, )>, W 1 nW 2 = <(i 1, i 1)>, W x =
<(!, 1,1, 1), (0, 0, 1, 1)>, W 2 = <(!, 1, h 1), (0, 1, 4, 3)), W x + W 2 =
<(i 1,1, 1), (0,1,4, 3), (0,0, 1, !)>.
IIIl
(1 2 3\ /l 2 3\
1,1 J , and identity permutation are even.
3. Of the 24 permutations, eight leave exactly one object fixed. They are per
mutations of three objects and have already been determined to be even.
Six leave exactly two objects fixed and they are odd. The identity permutation
is even. Of the remaining nine permutations, six permute the objects cyclically
and three interchange two pairs of objects. These last three are even since they
involve two interchanges.
4. Of the nine permutations that leave no object fixed, six are of one parity and
three others are of one parity (possibly the same parity). Of the 15 already
considered, nine are known to be even and six are known to be odd. Since
half the 24 permutations must be even, the six cyclic permutations must be
odd and the remaining even.
5. 77 is odd.
330
Answers to Selected Exercises
III2
1. Since for any permutation n not the identity there is at least one / such that
tt(/) < /, all terms of the determinant but one vanish. det A\ = Y\.l=x a a
2. 6.
4. (a) 32, (b) 18.
III3
1. 145; 134.
2. 114.
4. By Theorem 3.1, A A = (det A)I. Thus det A det A = (det A) n . If det A?±0,
then det A = (det A)"' 1 . If det .4=0, then A ■ A = 0. By (3.5), ^ a«^w =
for each k. This means the columns of A are linearly dependent and det A =
= (det A) n ~\
5. If det A 5* 0, then /? 1 = A/det A and /? is nonsingular. If det A = 0, then
det A = and /? is singular.
2/i
= I^(D n+i  1 det J B i ,
1=1
where 6 4 is the matrix obtained by crossing out row / and the last column,
n I n \
i=i u=i J
where Q, is the matrix obtained by crossing out column y and the last row of B t ,
= i i^(i) 2n+ ^ 3 (i) z+ Mo
«'=ii=i
n n
= I IM^= r T ^.
i=i j+i
III4
2.
3.
—X 3
— X 3
+ 2x 2
+ 6x 2
+ 5a:
 11a
 6.
+ 6.
'0

6"
4
1
1
*T.
1

2
1 
3
5. If A 2 + A + I = 0, then A(A  I) = I so that A  I = A~
Answers to Selected Exercises
331
6. If A is a real 3x3 matrix its characteristic polynomial/^) is real of degree 3. If
A satisfies a; 2 + 1 = 0, the minimum polynomial would divide x 2 + 1 and could
not have as an irreducible factor the real factor of degree one which /(x) must
have.
~~ 8. x 2 — 81 is the minimum polynomial.
1 1 .
III5
1. If I is an eigenvector with eigenvalue 0, then I is a nonzero vector in the
kernel of a. Conversely, if a is singular, then any nonzero vector in the kernel
of a is an eigenvector with eigenvalue 0.
2. If <r(!) = A£, then o a (f) = <r(A) = A 2 f. Generally, cr»(f) = A w .
3. If <r(f) = A x  and r(f) = A 2 , then (<r + t)() = <r(f) + t() = A x f + A 2  =
(A x + A 2 )£. Also, (fl<r)(f) = flff(f) = fl^f.
"1 0"
and
"2 2'
_2 2.
_o i_
4. Consider, for example,
5. o(A) is an eigenvalue of p(a) corresponding to f.
6. If <r(f) = A, then £ = ff 1 ^) = ^~ X (D so that f is an eigenvector of a x
corresponding to A 1 .
7. Let {Ij, . . . , !„} be a linearly independent set of eigenvectors with eigenvalues
{A l5 . . . , Aj. Then *(£, ^) = J, <r(S<) = £< ^i But since J, f< is also an
eigenvector (by assumption), we have <x(2t £*) = ^ 2* £*"• Tnus .2* (**' ~
A)£ f = 0. Since the set is linearly independent, A;  A = for each i.
8. x" is the characteristic polynomial and also the minimum polynomial. An
eigenvector p(x) must satisfy the equation D(p(x)) = kp(x). The constants
are the eigenvectors of D.
9. c is the corresponding eigenvalue.
11. If l x + £ 2 is an eigenvector with eigenvalue A, then A( x + l 2 ) = a (h + £2) =
Vi + Va Since (A  A^ + (A  A 2 )£ 2 = Owe have A  A x = A  A 2 = 0.
12. If £ = 2<=i fl <£f is an eigenvector with eigenvalue A, then A^^a^ = A£ =
*(£) = 2J=i«< ff (W = 2<=i W< Then SUW*  W< = and a t {X 
A 2 ) = for each i. Since the A t are distinct at most one of the A  A, can be zero.
For the other terms we must have a< = 0. Since f is an eigenvector, and hence
nonzero, not all a t = 0.
III6
1. 2, (1, 1); 7, (4, 5). 2. 3 + 2i, (1,0; 3  2i, (1_, /).
3. 3, (1, 2); 2, (2, 1). 4. 2, (^2, 1); 3, (1, Jl).
5. 4, (1,0, 0); 2, (3, 2, 0); 7, (24, 8, 9).
6. 1, (1,0,1); 2,(2, 1,0); 3,(0,1, 1).
7. 9,(4,1, 1); 9,(1, 4,0), (1,0,4).
8. l,(i\l,0); 3, (1,1,0), (0,0, 1).
III7
1. For each matrix A, P has the components of the eigenvectors of A in its
columns. Every matrix in the exercises of Section 6 can be diagonalized.
2. The minimum polynomial is (x — l) 2 .
332 Answers to Selected Exercises
3. Let a be the corresponding linear transformation. Since a ^ 1, there is a
nonzero vector £ x such that a(^ == £ 2 ^ ^.  2 ^ since cr is nonsingular.
If {£i> £2} is a basis, then the matrix representing a with respect to this basis
has the desired form. On the other hand, suppose that for every choice of
£i> (£i> £2} i s dependent. Then £ 2 is a multiple of £ x and every vector is an
eigenvector. By Exercise 7 of Section 5 this is impossible.
4. A~\AE)A = BA.
5. If ttj and v 2 are projections of the same rank k, we are asked to find a non
singular linear transformation a such that <r*ir x o = tt 2 . Let {a l5 . . . , a„} be a
basis such that ^(a^) = «.. for / <; k and ^(a^) = for / > k. Let {fa, ... ,
/?„} be a basis having similar properties with respect to tt 2 . Define a by the
rule o(fa) = a,. Then a^n^fa) = a^ir^a,) = <r\et t ) = fa for 1 < k, and
f^iPi) = ff 1 ^i(«i) = ff 1 (0) = for 1 > k. Thus a^^a = tt 2 .
6. By (7.2) of Chapter II, TrC^iM) = TriBAA' 1 ) = Tr(5).
IV1
1. (a) [1 1 1]; (c)[V2 0]; (</) [$ 1 0]. (b) and (<?) are not linear
functionals.
2. («){[1 0], [0 1 0], [0 1]}; (6){[1 1 0], [0 1 1],
[0 1]}; (c){[i i *],[* * *],[* I i».
5. If a = 2? =1 *<«<, then ^(a) = ^Xi *<M a i) = «y.
6. Let A = {a x , . . . , a„} be a basis such that 04 = a and a 2 = /?. Let >\ =
tA> • • • » 0n} be the dual basis. Then ^(a) = 1 and ^ X (/S) = 0.
7. Let /?(x) = x. Then <*<,(») # <* 6 0«).
8. The space of linear functionals obtained in this way is of dimension 1 , and
P n is of dimension n > 1.
9 /» = 2£i fn;=i (*  «*>) = 2u **<*)• /'k> = w.
10. ^CQU ****(*)) = 2*i W**(*» = 2U**y7j£) "„<**<*)) = Ay If
2*=x bkhii( x ) = 0» then ^(0) = 6, = 0. Thus {/*i(a), . . . , /*„(#)} is linearly inde
pendent. Since Oi{hj(x)) = 8 ijf the set {cr^ . . . , a n } is a basis in the dual space.
11. By Exercise 5,p(x) = 2*=i ^(pO*))*^*) = 2iU 7^ **(*)•
12. Let {a l5 . . . , a r } be a basis of W. Since <x ^ W, {a x , . . . , a r , a } is linearly
independent. Extend this set to a basis {a 1? . . . , a n } where a r+1 = a . Let
{<f> lt . . . , <£ n } be the dual basis. Then <f> r+1 has the desired property.
13. Let A = (aj, . . . , oc n } be a basis of V such that {a l5 . . . , a r } is a basis of W.
Let A ={#!,... , <£ w } be the basis dual to A. Let y = 2X=i v( a j)^j Then
for each 0^ e W, y> (a t ) = y(aj). Thus y and y> coincide on all of VV.
14. The argument given for Exercise 12 works for W = {0}. Since a ^ VV, there
is a <f> such that <£(a) 5* 0.
15. Let W = <j3>. If a £ W, by Exercise 12 there is a ^ such that <£(°0 = 1 and
*(/*) = 0.
IV2
1. This is the dual of Exercise 5 of Section 1.
2. Dual of Exercise 6 of Section 1. 3. Dual of Exercise 12 of Section 1.
4. Dual of Exercise 14 of Section 1. 5. Dual of Exercise 15 of Section 1.
Answers to Selected Exercises
333
IV3
1. P = (P 1 )^ = (PT)~\
1 1 o 
2. P =
1
1
1 1
(pl)T =
1
1 1
1 1 1
Thus,!' = {[1 1 1], [2 1 1], [1 1]}.
3. {[1 1 0], [0 1 1], [0 1]}.
4 {.L^j J ~2li \~l 2 — ' 2 J' L2 2 2]/"
5. BX = B(PX') = (BP)X' = B'X'.
IV4
1. («){[1 1 1]} (b){[\ 1 1 1]}.
2. [1 1 1].
3. Let W be the space spanned by {a}. Since dim W = 1, dim VVL = dim V 
1 <dimV. Thus there is a <£<£WL.
4. If <f> e TL, then <£a = for all aeT. But since S <= T, <£a = for all a e S
also. Thus TL «= SJ.
5. Since SLL is a subspace containing S, <S> <= SLL. By Exercise 4, S <= <S>,
SL => <S>^, SLJ c (S)ii = <s>.
6. Since S c S + T, SL => (S + T)L. Similarly, TL => (S + T)L. Thus (S +
T)± cjin TL. Since S nT c S, (S n T)i => SL. Similarly, (S n T)L =>
TL. Since (S n T)J is a subspace, (S n T)J = SL + TL.
7. If S and T are subspaces, SLJ = S and T^L = T. Thus SnT = JH n
TLL => (S^ + TL) J and hence (S n T)± <= (SL + TJ)LL = (SL + T±).
Similarly, S + T = S^i +THc(Si n TL)^ and hence (S + T)L =>
(SL n T±)L± = SL n TL.
8. SL + TL = (S n T)L = {0}^ = K.
9. SL n TL = (S + T)L = VJ = {0}.
10. By Exercises 9 and 10, SL + TL = V, and the sum is direct. For each y>eV
define y»i G S by the rule: v>i a = V a for all a e S. The mapping of y> G V onto
y x G S is linear and the kernel is SL. By Exercise 13 of Section 1 every functional
on S can be obtained in this way. Since V = SL © TL, S is isomorphic to TL.
11. f(t) = 4>(ta. + (1 /)/5) is a continuous function off. Since £(/ a + (1_ 00) =
/0(a) + (1  t)4>(P), fit) > if a, j8 e S+ and < t < 1. Thus a/3 c S+. If
a g S+ and /J G S~, then/(0) < and/(l) > 0. Since/is continuous, there is a
t, < t < 1, such that/(0 = 0.
IV5
A.
1. Let t be a mapping of U into V and er a mapping of V into W. Then, if <f> G W,
we have for all f eU, (^))(l) = ^(1)] = M^(l)] = a(*)[r(f)] =
2. {[6 2 1]}.
3. [1 2 1].
334
Answers to Selected Exercises
IV8
1.
1
1
"1 3 51
3 5 7
+
15 7 9_
and d&tA T = det {A) = (■
1
2
l) n det A
1 2
1
1 0_
Thus det A =
5. det A? = det A
det A.
7. <r,(a) = if and only if o f (*)(p) = /(<x, 0) = for all £ V.
9. Let dim U = m; dim V = «. /(a, #) = for all £ V means a e [^(V)]^ or,
equivalently, a e ct^KO). Thus p(r f ) = dim ^(V) = m — dim [^(V)] 1  =
m — dim ffKO) = m — v(o f ) = p(a f ).
10. If m 9* n, then either p(oy) < m or ,0(7^) < «.
11. L/ is the kernel of a f and V is the kernel of t / .
12. m — dim U = m — v(a f ) = p(o f ) = p(r f ) = n — v(r f ) = n — dim V .
13. =/(a + p, a + 0) = /(a, a) +/(a, 0) + /(/?, a) + /(£, /?) =/(«, /?) +
/(£, a).
14. If /4fi = BA, then C45) T = (BA) T = A T B T = AB.
15. B is skewsymmetric.
16. (a) (A 2 )? = A T A T = (AKA) = A 2 ;
(b) (AB  BA) T = (AB) T  (BA)T = B(A)  (A)B = AB  BA;
(c) (AB) T = (BA) T = A T B T = (~A)B = AB. If (AB)T = AB, then
AB = (AB) T = B T A T = B(A) = BA.
IV9
1. (a)
"2 f "
(c)
"1 1
2]
1 3
i
2
2 A
l_Z 2
7J
(«)
"1
2
r
2
4
2
1
2
1.
2. (a) 2x^2 + 1^2/2 + f#i# 2 + 6^2/2 (if ( x i, Vi) and (x 2 , ?/ 2 ) are the coordinates
of the two points), (c) x x x 2 + x x y 2 + x 2 y 1 + 2x x z 2 + lx 2 z x + 3y x y 2 + \y& 2 +
\y&\ + 7z 1 z 2 , (e) x x x 2 + 2x x y 2 + 2x 2 y x + 4y x y 2 + x x z 2 + x 2 z x + z x z 2 + 2y x z 2 +
1y 2 z x .
IV10
1. (In this and the following exercises the matrix of transition P, the order of
the elements in the main diagonal of P T BP, and their values, which may be
multiplied by perfect squares, are not unique. The following answers can only
T 2 2"
be thought of as representative possibilities.) (a) P =
2
U
The diagonal of P T BP is {1, 3, 9}; (b)
ri
2
1
Lo
2
7
4J
{1, 4,68};
(c)
"0
1
1
1
2
1
2
1
1
,{1,4, 1, 4}.
Answers to Selected Exercises
335
2. (a) P =
1 3"
4
, {2, 78} (c)
1
1
11"
1
3
.0
4
,0,2,30};
to
1
2
r
1
i_
0,0,0}.
IV11
1. (a)r=2,S = 2; (6)2,0; (c)3,3; (d)2,0; (e)l,l; (/) 2, 0; (,§03,1.
2. IfP =
1 b\2
L0 a J
, then P T BP =
a
_0 (a/4)(6 2 +4ac)J
3. There is a nonsingular g such that Q T AQ = I. Take P = Q~ .
4 Let P = /4 — 1
5* There is a nonsingular Q such that Q T AQ = B has r l's along the main
6. £f°£i ? = to,? T.' *>, r*r = 2«^ * "• ™» ™* =
(AX) T (AX) = Y T Y > for all real Z = (x lt . . . , x n ).
7 If y = (2/1, • • • ,y n ) * 0,then 7^7 > 0. HA * 0, there is an X = (x lt . . . ,x n )
such that AX = r ?* 0(why?). Then we would have = X T A T AX = Y T Y >
9. If ^JT * for any i, then = X^U ^ x = 25i X**?** > ° Thus
,4 ,jjf = for all X and ,4, = 0.
rv12
1. (a) P 
.0 U
.diagonal ={1,0} (b)
1 1 + f
L0
1 J
,{!,!}•
39. Proofs are similar to those for Exercises 39 of Section 11.
10. Similar to Exercise 14 of Section 8.
Vl
1. 6. 2. 2i.
6. (a  0, a + P) = (a, a)  (/S, a) + (a, /?)  (0, 0) = la 2 
7.  a + /5 2 =  a 2 +2(a,/3) + /S 2 .
11. (4 (0, 1, >. 0), HO, 2, 2, 1), K3, 2, 2, 8)j.
336 Answers to Selected Exercises
12. x* \,x*  3a>/5.
13. (a) If 2r=i a & = °> then J^ «,,(*„ *,) = (f„ ^ <,,*,) = (^, ) = for
each /. Thus 25i#« fl * = and the columns of G are dependent. (b) If
7L%xg<&i = for each /, then = J™ t «,(£, £) = (f„ £« x a^) for each i.
Hence, 2f=i^(^,2r=i«^) = (If=i^ 2r=i«^) =0. Thus ^^ =
0. (c) Let A = { ai , . . . , a n } be orthonormal, f, = 2"=i«*j a r Then ^„ =
(ft. f>) = ^k=i^ki a ki Thus G = ^*/i where ^ = [a i? ].
V2
1. If a = 2?=l a &> then (£<, a) = J?! «*(**, *i) = «i
X is linearly independen
Since (£,, /S) = 0, /? =
2. X is linearly independent. Let a £ V and consider = a — V" ^ " „ *' .
Since ($.. B) = 0.5 = 0. Z * =1 ll*<ll*
V4
1. ((^)*(a), /?) = (a, ardS)) = (**(a), r(/0) = (r*ff*(a), /?).
2. (<r(a), cr(a)) = (a, <7*cr(«)) = for all a.
3. (ff*(a), /?) = (a, a(0)) =/(a, 0) = /(/?, a) = (/?, a(a)) = _(*(«), 0) =
5. Let f be an eigenvector corresponding to A. Then A(, £) = (£, <x(f)) =
0*(£>, f) = (Af, I) = *(f, 5). Thus (A + A) = 0.
6. cr is skewsymmetric. 7. c is skewsymmetric.
8. Let $ e W^. Then for all r\ e W, (<r*(f), *?) = (£, <t(jj)) = 0.
10. Since (tt*) 2 = (tt 2 )* = n*, n* is a projection. £ e ^(tt*) if and only if
("*(£), *?) = (f , "(»?)) = for all *?; that is, if and only if I e SL. Finally,
(tt*(£), »?) = (I, *■(*?)) vanishes for all £ if and only if Tr(rj) = 0; that is, if and
only if r\ e T. Then tt*(V) £ TL. Since tt*(V) and TL have the same dimension,
11. (£, or(»?)) = (<r*(f), 17) = for all tj if and only if o*($) = 0, or £ £ W^.
13. By Theorem 4.3, V=W© W^. By Exercise 11, a*(V) = a*(W).
14. <7*(V) = ff*(W) = a*a(V). a(V) = aa*(V) is the dual statement.
15. a*(V) = a*a(V) = aa*(V) = a(V).
16. By Exercises 15 and 11, W 1  is the kernel of a* and a.
21. By Exercise 15, a(V) = a*(V). Then <r 2 (V0 = aa*{V) = a(V) by Exercise 14.
V5
1. Let I be the corresponding eigenvector. Then (£, £) = (<r(l), CT (f)) = 0*f» ^£)
AA(£, I).
3. It also maps £ 2 onto ±
*i"*.
V2
4. For example, f 2 ont ° K 2 ^i ~~ 2 ^2 + h) anc * £3 ont o i(2fi + £ 2 — 2 ^)
Answers to Selected Exercises 337
V6
1. (a) and (c) are orthogonal. 2. (a).
5. (a) Reflection in a plane (x x , x 2 plane). (b) 180° rotation about an axis (x 3 axis).
(c) Inversion with respect to the origin, (d) Rotation through about an axis
(x 3 axis). (e) Rotation through about an axis (x 3 axis) and reflection in the
perpendicular plane (x lf x 2 plane). The characteristic equation of a thirdorder
orthogonal matrix either has three real roots (the identity and (a), (b), and (c)
represent all possibilities) or two complex roots and one real root ((d) and (e)
represent these possibilities).
V7
1. Change basis in V as in obtaining the Hermite normal form. Apply the Gram
Schmidt process to this basis.
2. If a( Vj ) = 2Li a i7Vi, then a*(r, k ) = J^ =1 (Vj, <**))>?; = SUMty), Vk)% =
2?=i (2i=v "aVi, nic)ni = 2U*«^
3. Choose an orthogonal basis such that the matrix representing a* is in super
diagonal form.
V8
1. (a) normal; (b) normal; (c) normal; (d) symmetric, orthogonal; (V) orthog
onal, skewsymmetric; (/) Hermitian; (g) orthogonal; (h) symmetric,
orthogonal; (/) skewsymmetric; normal; (j) nonnormal ; (&) skewsymmetric
normal.
2. All but (c) and (/). 3. AT A = (A)A = A 2 = AA?.
"0 1 r
5.
V9
1 1
/ 1
6. Exercise 1(c).
4. (<r*(°0, /S) = (a, o(fi)) =/(a, /?) =/(/?, /?) = (£, a(a)) = (<7(a), 0).
5. /(a, 0) = (a, a(fi)) = (£? =1 a^, JjU V(£)) = Q> =1 a, £„ £? =1 W>) =
iLi MA
6. ?(a) =/(a, a) = £? =1 kl 2 A,, Since £ i=1 k 2 = 1, min {AJ < a(a) <
max {A,} for a e S, and both equalities occur. If a ^ 0, there is a real positive
scalar a such that ao. e S. Then o(<x) = ^ ^(aa) > min {Aj > 0, if all eigen
values are > 0. a
V10
1. (a) unitary, diagonal is {1, /}. (b) Hermitian, {2, 0}. (c) orthogonal, {cos +
/sin0, cos — /sin 0}, where = arccos 0.6. (d) Hermitian, {1,4}.
(e) Hermitian, {1,1+ yfl, 1  yfl).
338 Answers to Selected Exercises
Vll
1. Diagonal is {15, 5}. (d) {9, 9, 9}. (e) {18, 9, 9}. (/) {9, 3, 6}.
(^{9,0,0}. (/*){1,2,0}. (i){l, 1, ft}. (/){3,3, 3}. (*){3,6,6}.
2. (</), (A).
3. Since P T BP = B' is symmetric, there is an orthogonal matrix R such that
ijrg'/? = B" is diagonal matrix. Let Q = PR. Then (PR) T A(PR) =
RT P T APR = /JTR = l an d (PR)T B ( PR ) = RT P T BPR = /?T£'/? = 5" j s
diagonal.
4. ^ T /4 is symmetric. Thus there is an orthogonal matrix Q such that Q T (A T A)Q =
D is diagonal. Let B = Q T AQ.
5. Let P be given as in Exercise 3. Then det {P T BP  xl) = det (P T (B  xI)P) =
detP 2 • det {B — xA). Since P T BP is symmetric, the solutions of det (B — xA) =
are also real.
[a bl
7. Let A = . Then A is normal if and only if b 2 = c 2 and ab + cd =
_c dj
ac + bd. If b or c is zero, then both are zero and .4 is symmetric. If a =0,
then b 2 = c 2 and cd = bd. If d ?* 0, c = d and /I is symmetric. If d = 0,
either 6 = c and /I is symmetric, or b = —c and A is skewsymmetric.
8. If b = c, A is symmetric. If b = —c, then a = d.
9. The first part is the same as Exercise 5 of Section 3. Since the eigenvalues of
a linear transformation must be in the field of scalars, o has only real eigen
values.
10. a 2 = — a* a is symmetric. Hence the solutions of \A 2 — xl\ = are all
real. Let A be an eigenvalue of a 2 corresponding to f. Then (<x(f),
<t(£)) = (£, <7*<t(!)) = (f, (7 2 (£)) = A(£, I). Thus * ^ 0. Let A = ^
a(*0 = 1 <r 2 (f) = I (^f) = /,!. cr 2 (r/) = /uo(Z) = fn (£, >?) =
/X [*
rj).
11. <x(£) = ^r?, ff(??) = /x£.
12. The eigenvalues of an isometry are of absolute value 1. If <?(£) = A with A
real, then a* (I) = AS, so that (<r + <r*)(£) = 2A£.
13. If (a + (T*)(£) = 2//£and<r(f) = A£,then A = ±1 and2^ = 2A = ±2. Since
(f , ct(I)) = (a*(£), ) = (, 0*(£)) 2MI, f) = (£, (* + **)(£)) = 2(f , *(£)).
Thus \n\ • Hill 2 = (f, <r(f)) < Hill • a(f) = Hill 2 , and hence /* < 1. If
\ju\ = 1, equality holds in Schwarz's inequality and this can occur if and only
if cx(£) is a multiple of £. Since £ is not an eigenvector, this is not possible.
14. (£, rj) = * {(£, <x())  Mf, I)} = 0. Since <x(£) + <r*(£) = 2/uf,
<r 2 (£) + f = 2/xo(S). Thus
, (T 2 ()  ixa^) fio($)   /, 2 £ + ^Vl  fx 2 v  f
ct(?j) = — = — = ,
Vi  ^ Vi  /« 2 Vi  ^ 2
=  y/l  n 2 + vy.
1 5. Let f x , ^ x be associated with /x lt and  2 , j? 2 be associated with /x 2 , where /x 2 ^ /x 2 .
Then (f x , (a + <r*)(£ 2 )) = (f lf 2/i 2 S 2 ) = ((a + cr*^^), f 2 ) = (2^!^, I 2 ).
Thus (f lf f a ) = 0.
Answers to Selected Exercises
V12
2. A + A* =
2
3
4
3
0"
4
3
2
3
2
3_
339
. The eigenvalues of A + A* are {— §, f , 2}.
Thus ii = —\ and 1. To n = 1 corresponds an eigenvector of /I which is
—= (1, — 1, 0). An eigenvector of A + A* corresponding to f is (0, 0, 1).
V2 1
If this represents f, the triple representing i\ is —= (1, 1,0). The matrix repre
V 2
senting or with respect to the basis {(0, 0, 1), —= (1, 1, 0), —= (1, 1, 0)} is
y 2 >/2
2V2
2V2
.
u
VI1
1. (1)(1,0,1) + ^(1,1,1) +/ a (2, 1,0); [1 2 l]^,^,^) =2.
(2) (1, 2, 2) + t x {2, 1, 2) + r 2 (2, 2, 1); [1 2 2](x 1} a a , * 3 ) = 9.
(3) (1,1, 1,2) +^(0,1,0, l)+/ 2 (2, 1, 2,3); [1 1 OK^,^,^,^) =
2, [2 1 1](%, cc 2 , x 3 , z 4 ) = 1.
2. (6, 2, 1) + t(6, 1, 4); [1 2 l](aj lf x t , x 3 ) = 2, [1 2 2](x x , x s ,
x ) = 9.
3. L 3 = (2,'l, 2) + <(0, 1, 1), (3, 0, f)>.
4. ^(1, 1) + ii(6, 7) + f(5, 6) = (0, 0).
6. Let L x and L 2 be linear manifolds. If L, n L 2 / 0, let <x e L x n L 2 . Then
L x = <x + S x and L 2 = a + S 2 , where S x and S 2 are subspaces. Then L x n
L 2 = a + (S x n S 2 ).
7. Clearly, a x + S x <= 04 + (a 2 — a x > + S x + S 2 and a 2 + S 2 = 04 + (a 2 — a x ) +
S 2 c aj + <a 2 — a x > + S x + S 2 . On the other hand, let 04 + S be the join
of /! and L 2 . Then L t = 04 + S x <= Kl + S implies S x <= s, and L 2 = <x 2 +
^2 c a i + ^ implies a 2 — a x + S 2 <= S. Since S is a subspace, <<x 2 — a x > + S x +
S 2 c= S. Since, a x + S is the smallest linear manifold containing L x and L 2 ,
04 + S = 04 + <a 2 — a x > + S x + S 2 .
8. If a g L x n L 2 , then L x = a + S x and L 2 = a + S 2 . Thus ^Ji^ = a +
<a — a > + S x + S 2 = a + S x + S 2 . Since 04 e ^Jl^, L X JL 2 = 04 + S x +
s 2 .
9. If a 2 — a x g Sj + S 2 , then a 2 — a x = ^ + fi 2 where ^ G S x and /? 2 G S 2 . Hence
a 2 — £2 = a i + Pi Since a x + fl x g a x + S x = L 2 and a 2 — /? 2 g a 2 + S 2 =
L 2 , L x n L 2 ^ 0.
10. If l x n L 2 # 0, then L X JL 2 = a x + S x + S 2 . Thus dim L^^ = dim (S x + S 2 ).
If ^013= 0, then L^iL^ = a x + <a 2 — a x > + S x + S 2 and L a JL 2 ^ a x +
S 2 + S 2 . Thus dim ^Jlg = dim (S x + S 2 ) + 1.
340 Answers to Selected Exercises
VI2
1. If Y = [yL y 2 y 3 ], then Y must satisfy the conditions Y(\, 1, 0) >
7(1,0, 1)>0, F(0, 1,1) >0
2. {[1 1 1], [1 1 1], [1 1 1]}.
3. {(1, 1, 0), (1, 0, 1), (0, 1, 1), (0, 1, 1), (1, 1, 0), (1, 1, 1)}.
4 {(1, 0, 1), (0, 1, 1), (0, 1, 1)}. Express the omitted generators in terms of
the elements of this set.
5. {[1 1 2], [1 1 1], [1 1 1]}.
6. {(1,0, 1), (3,1,2), (1, 1,0)}.
7. Let Y = [1 1 2]. Since YA > and YB = 2 < 0, (1, 1, 0)<£ C 2 .
8. Let Y = [2 2 1]. 9. Let Y = [1 2].
10. This is the dual of Theorem 2.14. 11. Let Y = [2 2 11.
12. Let A = {<f> x , ... , <f> n } be the dual basis to A. Let <f> = ]>> =1 <t>i Then £ is
semipositive if and only if f > and <£ f > 0. In Theorem 2.11, take /S =
and g = 1. Then v>£ = <g = 1 for all ?»6^ and the last condition in
(2) of Theorem 2.11 need not be stated. Then the stated theorem follows
immediately from Theorem 2.11.
14. Using the notation of Exercise 13, either (1) there is a semipositive £ such
that <r(£) = 0, that is, £ e W, or (2) there is a y> E V such that 6{y>) > 0. Let
<f> = d(y>). For £eW, <t>£ = d(y>)! = vM£) = 0. Thus <f> e W±.
15. Take £ = 0, g = 1, and <f> = £f =1 &, where {^ . . . , <£„} is the basis of P±.
VI3
1. Given A, B, C, the primal problem is to find X > which maximizes CJSf
subject to AX <: B. The dual problem is to find Y > which minimizes
r£ subject to YA <, C.
2. Given A, B, C, the primal problem is to find X > which maximizes CX
subject to AX = B. The dual problem is to find Y which minimizes YB
subject to Y/l > C.
6. The pivot operation uses only the arithmetic operations permitted by the
field axioms. Thus no tableau can contain any numbers not in any field
containing the numbers in the original tableau.
7. Examining Equation (3.7) we see that «£f will be smaller than <f>£ if c k — d k <
0. This requires a change in the first selection rule. The second selection
rule is imposed so that the new £' will be feasible, so this rule should not be
changed. The remaining steps constitute the pivot operation and merely
carry out the decisions made in the first and second steps.
8. Start with the equations
Ax x + x 2 + x 4 =10
The first feasible solution is (0, 0, 6, 10, 3). The optimal solution is (2, 2, 0,
0, 3). The numbers in the indicator row of the last tableau are (0, 0, — f , — , 0).
9. The last three elements of the indicator row of the previous exercise give
Vi = f , 2/ 2 = h 2/3 = °
Answers to Selected Exercises
341
10. The problem is to minimize 6y x + I0y 2 + 3y 3 + My 4 + My 5 , where M is
very large, subject to
2/i + 2/2 + 2/3 + 2/4  2/e = 2
22/1 + 4y 2  y 3 +y 5  2/7 = 5.
When the last tableau is obtained, the row of {</*} will be [6 10 2 2
—2 —2]. The fourth and fifth elements correspond to the unit matrix in
the original tableau and give the solution x x = 2, x 2 = 2 to Exercise 8.
11. Maximum = 12 at x x = 2, x 2 = 5.
12. (0,0), (0,2), (1,4), (2, 5).
15. Xand Ymect the test for optimality given in Exercise 14, and both are optimal.
16. AX = B has a nonnegative solution if and only if min FZ = 0.
. Vi 6 4 > "» 164) 164 v/» 164/.
VI6
1. (a) A = (1)
L 2
1
2J
+ 3
(b) A = 2
(c) A = 5
(*M = 3
5J
2.1
5
+ (3)
A =2
+ (7)
2. ^ =
"2
7
3
11
14
6
U10
r j_
5
_2_
5
1 J
_2_
5
+ (8)
1
3
L 10
__3_~1
10
+ 2
4. /4 = ^.E^ + AE 2 where E x
18
6
3
3 3
2 2
2 3
3
42"
14
7
"I V 311
+ 3
_0
0_
0"
■2 3
, E 2 =
2 3
1_
2 3
and ^Fi = 2E X + N x , where N x =
6. e^ = e 2
3
2
1 1
3
2
+ e
2
1
2 3
2 3
2 3
6
4
2
0"
3"
2
1
VI8
1. V = \(x 2
2/i)
*i) 2 +
y/3
2(^2 — ^3) ^~ (2/2 —
+
/3
iC^s ~ x i) +^r (2/3
342
Answers to Selected Exercises
2. £/.
2
4. These displacements represent translations of the molecule in the plane con
taining it. They do not distort the molecule, do not store potential energy,
and do not lead to vibrations of the system.
VI9
2. tt = (124), a = (234), an = (134), P = an' 1 = (12)(34).
3. Since the subgroup is always one of its cosets, the alternating group has only
two cosets in the full symmetric group, itself and the remaining elements.
Since this is true for both right and left cosets, its right and left cosets are equal.
(<?) = Z>((123)) = £((132)) = [1],£((12)), = £((13)) = £((23)) = [1].
" 2 r
7. The matrix appearing in (9.7) is H = 4
10.
11.
is then P =
2V6
"3 1"
2
1
The matrix of transition
G is commutative if and only if every element is conjugate only to itself. By
Theorem 9.11 and Equation 9.30, each n r = 1.
Let £ = 6>2in7n be a primitive «th root of unity. If a is a generator of the cyclic
group, let £*(a) = [£*], k = 0, ...,« 1.
c 4
D 1
1
1
1
1
D 1
1
1
1
1
£ 2
1
i
1
— /
£ 2
1
1
1
1
£ 3
1
1
1
1
£ 3
1
1
1
1
£ 4
1
— /
1
i
£ 4
1
1
1
1
12. By Theorem 9.12 each n r \p 2 . But n r = p or p 2 is impossible because of (9.30)
and the fact that there is at least one representation of dimension 1. Thus
each n r = 1 , and the group is commutative.
16. Since ab must be of order 1 or 2, we have (ab) 2 = e, or ab = b  ^ 1 . Since
a and b are of order 1 or 2, a 1 = a and b _1 = b.
17. If G is cyclic, let a be a generator of G, let £ = e™'*, and define £ fc (a) = [£*].
If G contains an element a of order 4 and no element of higher order, then G
contains an element b which is not a power of a. b is of order 2 or 4. If b is of
order 4, then b 2 is of order 2. If b 2 is a power of a, then b 2 = a 2 . Then c = ab
is of order 2 and not a power of a. In any event there is an element c of order 2
which is not a power of a. Then G is generated by a and c. If G contains
elements of order 2 and no higher, let a, b, c be three distinct elements of order 2.
They generate the group. Hints for obtaining the character tables for these last
two groups are given in Exercises 21 , 25, and 26.
18. The character tables for these two nonisomorphic groups are identical.
29. I 2 + l 2 + 2 2 + 3 2 + 3 2 .
30. H 4 contains C x (the conjugate class containing only the identity), C 3 (the
class containing the eight 3cycles), and C 5 (the class containing the three pairs
of interchanges).
Answers to Selected Exercises
343
VI10
2. The permutation (123) is represented by
_1
2
V3/:
V3/2 i
i 
n/3/2
V3/2
*
*
V3/2
sentation
of (12)
V3/2
is
*
~
1
0~
1
1
1

1
1
3. Ci — 1,^2 — 1*^3 — 2.
4. ^ = (V3/2, i,V3/2, 4,0,1).
The displacement is a uniform expansion of the molecule.
5 £ 2 = (,f, h f,l,0).
This displacement is a rotation of the molecule without storing potential energy
6. {£, = (1, 0, 1, 0, 1, 0),  4 = (0, 1, 0, 1, 0, 1)}. This subspace consists of
translations without distortion in the plane containing the molecule.
7. This subspace is spanned by the vectors £ 5 and f 6 given in Exercise 8.
Notation
S„:/*eM
{aP}
<A>
+ (for sets)
Im(<r)
Hom(a, I/)
K(<x)
v(<t)
U/K
Sgn 7T
det^
l fl y
adj A
C(k)
S{X)
TiU)
V (space)
A (basis)
W 1
R
6
A x B
147
6
)Q = i A
147
6
y a
147
6
©7= i^i
148
12
150
21
fs
159
23
Jss
159
27
A
171
28
A*
171
30
(a,0
177
31
II «i
111
31
d{<x, P)
111
31
rj
186
55
a*
189
80
W t 1 W 2
191
84
A(S)
225
87
H(S)
228
89
LiJL 2
229
89
W +
230
95
P (positive orthant)
234
106
> (for vectors)
234
107
> (for vectors)
238
115
/'(& n)
262
129
df^)
263
130
e A
275
139,191
X (for ntuples)
278
140
D(G) (representation)
294
142
1
298
345
Index
Abelian group, 8, 293
Addition, of linear transformations, 29
of matrices, 39
of vectors, 7
Adjoint, of a linear transformation, 189
of a system of differential equations, 283
Adjunct, 95
Affine, closure, 225
combination, 224
nspace, 9
Affinely dependent, 224
Algebraically closed, 106
Algebraic multiplicity, 107
Alternating group, 308
Annihilator, 139, 191
Associate, 76
Associated, homogeneous problem, 64
linear transformation, 192
Associative algebra, 30
Augmented matrix, 64
Automorphism, 46, 293
inner, 293
Basic feasible vector, 243
Basis, 15
dual, 130
standard, 69
Bessel's inequality, 183
Betweenness, 227
Bilinear form, 156
Bounded linear transformation, 260
Cancellation, 34
Canonical, dual linear programming
problem, 243
linear programming problem, 242
mapping, 79
Change of basis, 50
Character, of a group, 298
table, 306
347
Characteristic, equation, 100
matrix, 99
polynomial, of a matrix, 100
of a linear transformation, 107
value, 106
Characterizing equations of a subspace, 69
Codimension, 139
Codomain, 28
Cofactor, 93
Column rank, 41
Commutative group, 8, 293
Companion matrix, 103
Complement, of a set, 5
of a subspace, 23
Complementary subspace, 23
Complete inverse image, 27
Completely reducible representation, 295
Complete orthonormal set, 183
Completing the square, 166
Component of a vector, 17
Cone, convex, 230
dual, 230
finite, 230
polar, 230
polyhedral, 231
reflexive, 231
Congruent matrices, 158
Conjugate, bilinear form, 171
class, 294
elements in a group, 294
linear, 171
space, 129
Continuously differentiable, 262, 265
Continuous vector function, 260
Contravariant vector, 137, 187
Convex, cone, 230
hull, 228
linear combination, 227
set, 227
Coordinate, function, 129
space, 9
348
Index
Coordinates of a vector, 17
Coset, 79
Covariant vector, 137, 187
Cramer's rule, 97
Degenerate linear programming problem,
246
Derivative, of a matrix, 280
of a vector function, 266
Determinant, 89
Vandermonde, 93
Diagonal, main, 38
matrix, 38, 113
Differentiable, 261, 262, 265
Differential of a vector function, 263
Dimension, of a representation, 294
of a vector space, 1 5
Direct product, 150
Direct sum, external, 148, 150
internal, 148
of representations, 296
of subspaces, 23, 24
Directional derivative, 264
Direct summand, 24
Discriminant of a quadratic form, 199
Distance, 177
Divergence, 267
Domain, 28
Dual, bases, 142
basis, 134
canonical linear programming problem,
243
cone, 230
space, 129
spaces, 134
standard linear programming problem,
240
Duality, 133
Eigenspace, 107
Eigenvalue, 104, 192
problem, 104
Eigenvector, 104, 192
Elementary, column operations, 57
matrices, 58
operations, 57
Elements of a matrix, 38
Empty set, 5
Endomorphism, 45
Epimorphism, 28
Equation, characteristic, 100
minimum, 100
Equations, linear, 63
linear differential, 278
standard system, 70
Equivalence, class, 75
relation, 74
Equivalent representations, 296
Euclidean space, 179
Even permutation, 87
Exact sequence, 147
Extreme vector, 252
Factor, group, 293
of a mapping, 81
space, 80
Faithful representation, 294
Feasible, linear programming problem,
241, 243
subset of a basis, 243
vector, 241, 243
Field, 5
Finite, cone, 230
dimensional space, 15
sampling theorem, 212
Flat, 220
Form, bilinear, 156
conjugate bilinear, 171
Hermitian, 171
linear, 129
quadratic, 160
Fourgroup, 309
Fourier coefficients, 182
Functional, linear, 129
Fundamental solution, 280
General solution, 64
Generators of a cone, 230
Geometric multiplicity, 107
Gradient, 136
Gramian, 182
GramSchmidt orthonormalization
process, 179
Group, 8, 292
abelian, 8, 293
alternating, 308
commutative, 8, 293
factor, 293
order of, 293
symmetric, 308
Index
349
Halfline, 230
HamiltonCay ley theorem, 100
Hermite normal form, 55
Hermitian, congruent, 172
form, 171
matrix, 171
quadratic form, 171
symmetric, 171
Homogeneous, associated problem, 64
Homomorphism, 27, 293
Hyperplane, 141, 220
Idempotent, 270
Identity, matrix, 46
permutation, 87
representation, 308
transformation, 29
Image, 27, 28
inverse, 27
Independence, linearly, 1 1
Index set, 5
Indicators, 249
Induced operation, 79
Injection, 146, 148
Inner, automorphism, 293
product, 177
Invariant, subgroup, 293
subspace, 104
under a group, 294
Inverse, image, 27
matrix, 46
transformation, 43
Inversion, of a permutation, 87
with respect to the origin, 37
Invertible, matrix, 46
transformation, 46
Irreducible representation, 271, 295
Isometry, 194
Isomorphic, 18
Isomorphism, 28, 293
Jacobian matrix, 266
Join, 229
Jordan normal form, 118
Kernel, 31
Kronecker delta, 15
Kronecker product, 310
Lagrange interpolation formula, 132
Lagrangian, 287
Length of a vector, 177
Line, 220
segment, 227
Linear, 1
algebra, 30
combination, 11
nonnegative, 230
conditions, 221
constraints, 239
dependence, 11
form, 129
functional, 129
independence, 11
manifold, 220
problem, 63
relation, 11
transformation, 27
Linearly, dependent, 11
independent, 11
Linear programming problem, 239
Linear transformation, 27
addition of, 29
matrix representing, 38
multiplication of, 30
normal, 203
scalar multiple of, 30
symmetric, 192
Main diagonal, 38
Manifold, linear, 220
Mapping, canonical, 29
into, 27
natural, 29
onto, 28
Matrix polynomial, 99
Matrix, 37
addition, 39
characteristic, 99
companion, 103
congruent, 158
diagonal, 38
Hermitian, 171
congruent, 172
identity, 46
normal, 201
of transition, 50
product, 40
representing, 38
scalar, 46
350
Index
Matrix (continued)
sum, 39
symmetric, 158
unit, 46
unitary, 194
Maximal independent set, 14
Mechanical quadrature, 256
Minimum, equation, 100
polynomial, 100
Monomorphism, 27
Multiplicity, algebraic, 107
geometric, 107
ndimensional coordinate space, 9
Nilpotent, 274
Nonnegative, linear combination, 230
semidefinite, Hermitian form, 168
quadratic form, 173
Nonsingular, linear transformation,
matrix, 46
Non trivial linear relation, 11
Norm of a vector, 177
Normal, coordinates, 287
form, 76
Hermite form, 55
Jordan form, 118
linear transformation, 203
matrix, 201
over the real field, 176
subgroup, 293
Normalized vector, 178
Normalizer, 294
Nullity, of a linear transformation, 31
of a matrix, 41
Objective function, 239
Odd permutation, 87
Onetoone mapping, 27
Onto mapping, 28
Optimal vector, 241
Order, of a determinant, 89
of a group, 293
of a matrix, 37
Orthant, positive, 234
Orthogonal, linear transformation, 270
matrix, 196
similar, 197
transformation, 194
vectors, 138, 178
Orthonormal, basis, 178
Parallel, 221
Parametric representation, 221
Parity of a permutation, 88
Parseval's identities, 183
Particular solution, 63
Partitioned matrix, 250
Permutation, 86
even, 87
identity, 87
group, 308
odd, 87
Phase space, 285
Pivot, element, 249
operation, 249
Plane, 220
Point, 220
Pointed cone, 230
Polar, 162
cone, 230
form, 161
Pole, 162
Polyhedral cone, 231
Polynomial, characteristic, 100
matrix, 99
minimum, 100
Positive, orthant, 234
vector, 238
Positivedefinit, Hermitian form, 173
quadratic form, 168
Primal linear programming problem, 240
Principal axes, 287
Problem, associated homogeneous, 64
eigenvalue, 104
linear, 63
Product set, 147
Projection, 35, 44, 149
Proper subspace, 20
Quadratic form, 160
Hermitian, 171
Quotient space, 80
Rank, column, 41
of a bilinear form, 164
of a Hermitian form, 173
of a linear transformation, 31
of a matrix, 41
row, 41
Real coordinate space, 9
Reciprocal basis, 188
Index
351
Reducible representation, 295
Reflection, 43
Reflexive, cone, 231
law, 74
space, 133
Regular representation, 301
Relation, of equivalence, 74
linear, 11
Representation, identity, 308
irreducible, 271, 295
of a bilinear form, 157
of a change of basis, 50
of a group, 294
of a Hermitian form, 171
of a linear functional, 130
of a linear transformation, 38
of a quadratic form, 161
of a vector, 18
parametric, 221
reducible, 295
Representative of a class, 75
Resolution of the identity, 271
Restriction, mapping, 84
of a mapping, 84
Rotation, 44
Rowechelon, form, 55
Sampling, function, 254
theorem, 253
Scalar, 7
matrix, 46
multiplication, of linear transformations,
30
of matrices, 39
of vectors, 7
product, 177
transformation, 29
Schur's lemma, 297
Schwarz's inequality, 177
Selfadjoint, linear transformation, 192
system of differential equations, 283
Semidefinite, Hermitian form, 173
quadratic form, 168
Semipositive vector, 238
Sgn, 87
Shear, 44
Signature, of a Hermitian form, 173
of a quadratic form, 168
Similar, linear transformations, 78
matrices, 52, 76
orthogonal, 197
unitary, 197
Simplex method, 248
Singular, 46
SkewHermitian, 193
Skewsymmetric, bilinear form, 158
linear transformation, 192
matrix, 159
Solution, fundamental, 280
general, 64
particular, 63
Space, Euclidean, 179
untary, 179
vector, 7
Span, 12
Spectral decomposition, 271
Spectrum, 270
Standard, basis, 69
dual linear programming problem, 240
primal linear programming problem, 239
Steinitz replacement theorem, 13
Straight line, 220
Subgroup, invarient, 293
Subspace, 20
invariant under a linear transformation,
104
invariant under a representation, 295
Sum of sets, 39
Superdiagonal form, 199
Sylvester's law of nullity, 37
Symmetric, bilinear form, 158
group, 308
Hermitian form, 192
law, 74
linear transformation, 192
matrix, 158
part of a bilinear form, 159
Symmetrization of a linear transformation,
295
Symmetry, of a geometric figure, 307
of a system, 312
Tableau, 248
Trace, 115,298
Transformation, identity, 29
inverse, 43
linear, 27
orthogonal, 194
scalar, 29
unit, 29
352
Index
Transformation (continued)
unitary, 194
Transition matrix, 50
Transitive law, 74
Transpose of a matrix, 55
Trivial linear relation, 11
Unitary, matrix, 196
similar, 197
space, 179
transformation, 194
Unit matrix, 46
Vandermonde determinant, 93
Vector, 7
feasible, 241, 243
normalized, 178
optimal, 241
positive, 238
semipositive, 238
space, 7
Vierergruppe {see Fourgroup), 309
Weierstrass approximation theorem, 185
Zero mapping, 28
X
'
NERING
LINEAR ALGEBRA AND
MATRIX THEORY
By EVAR D, NERING, Professor of Mathematics,
Arizona State University
"The author presents Ideas in linear algebra very effectively with the help
of matrices. ...The introductions are... excellent and help clarify the
materia] substantially. The discussions in the introduction as welJ as in
the body of each chapter are also very illuminating in details This
book is highly recommended as a textbook. 11 — Physics Today on the first
edition of Nering.
The major change in the second edition is in the addition of new material.
More sophisticated mathematical material is included and may be used
independently by the reader with advanced knowledge of linear algebra.
A new section on applications is added which provides an introduction to
the modern treatment of calculus of several variables. The concept of
duality receives expanded treatment in this edition. Finally, the appendix
supplies a "doityourself" kit which allows the reader to make up any
number of exercises from those in the book.
The presentation of material is handled with greater clarity and precision.
The author has made changes in terminology regarding characteristic
equations and normal linear transformations in order to clarify the es
sential mathematical ideas. Notations have been changed to correspond
to more current usage.
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